Question 13 Marks
If $y=A e^{m x}+B e^{n x}$, show that $y_2-(m+n) y_1+(m n) y=0$.
View full question & answer→Question 23 Marks
If $y=f(x)$ is a differentiable function of $x$, then show that $\frac{d^2 x}{d y^2}=-\left(\frac{d y}{d x}\right)^{-3} \cdot \frac{d^2 y}{d x^2}$
View full question & answer→Question 33 Marks
If $x =e^{\frac{x}{y}}$, then show that $\frac{d y}{d x}=\frac{x-y}{x \log x}$
View full question & answer→Question 43 Marks
If $\sin y=x \sin (a+y)$, then show that $\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}$
View full question & answer→Question 53 Marks
Differentiate the following w.r.t. x:$\tan ^{-1}\left[\frac{\sqrt{x}(3-x)}{1-3 x}\right]$
View full question & answer→Question 63 Marks
Differentiate the following w.r.t. x:$\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$
View full question & answer→Question 73 Marks
Differentiate the following w. r. t. x.$y=\log \left[\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}-x}\right]$
Answer$
\begin{aligned}
y=\log \left[\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}-x}\right] & =\log \left[\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}-x} \times \frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}+x}\right] \\
& =\log \left[\frac{\left(\sqrt{x^2+a^2}+x\right)^2}{x^2+a^2-x^2}\right] \\
& =\log \left[\frac{\left(\sqrt{x^2+a^2}+x\right)^2}{a^2}\right] \\
& =\log \left(\sqrt{x^2+a^2}+x\right)^2-\log \left(a^2\right) \\
\therefore \quad & =2 \log \left(\sqrt{x^2+a^2}+x\right)-\log \left(a^2\right)
\end{aligned}
$
Differentiate $w, r, t . x$$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[2 \log \left(\sqrt{x^2+a^2}+x\right)-\log \left(a^2\right)\right] \\ & =2 \frac{d}{d x}\left[\log \left(\sqrt{x^2+a^2}+x\right)\right]-\frac{d}{d x}\left[\log \left(a^2\right)\right] \\ & =2 \times \frac{1}{\sqrt{x^2+a^2}+x} \cdot \frac{d}{d x}\left[\sqrt{x^2+a^2}+x\right]-0 \\ & =\frac{2}{\sqrt{x^2+a^2}+x} \cdot\left[\frac{1}{2 \sqrt{x^2+a^2}} \cdot \frac{d}{d x}\left(x^2+a^2\right)+1\right] \\ & =\frac{2}{\sqrt{x^2+a^2}+x} \cdot\left[\frac{1}{2 \sqrt{x^2+a^2}}(2 x)+1\right] \\ & =\frac{2}{\sqrt{x^2+a^2}+x} \cdot\left[\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\right] \\ \frac{d y}{d x} & =\frac{2}{\sqrt{x^2+a^2}}\end{aligned}$
View full question & answer→Question 83 Marks
Differentiate the following w. r. t. x.$y=\log _3\left(\log _5 x\right)$
View full question & answer→Question 93 Marks
Find the $n^{\text {th }}$ derivative of the following : $\sin x$
AnswerLet $y=\sin x$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}(\sin x)=\cos x \\
& \frac{d y}{d x}=\sin \left(\frac{\pi}{2}+x\right)
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left[\sin \left(\frac{\pi}{2}+x\right)\right] \\
& \frac{d^2 y}{d x^2}=\cos \left(\frac{\pi}{2}+x\right) \frac{d}{d x}\left(\frac{\pi}{2}+x\right) \\
& \frac{d^2 y}{d x^2}=\sin \left(\frac{\pi}{2}+\frac{\pi}{2}+x\right)(1) \\
& \frac{d^2 y}{d x^2}=\sin \left(\frac{2 \pi}{2}+x\right)
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=\frac{d}{d x}\left[\sin \left(\frac{2 \pi}{2}+x\right)\right] \\
& \frac{d^3 y}{d x^3}=\cos \left(\frac{2 \pi}{2}+x\right) \frac{d}{d x}\left(\frac{2 \pi}{2}+x\right) \\
& \quad=\sin \left(\frac{\pi}{2}+\frac{2 \pi}{2}+x\right)(1) \\
& \frac{d^3 y}{d x^3}=\sin \left(\frac{3 \pi}{2}+x\right)
\end{aligned}
$
In general $n^{\text {th }}$ order derivative will be
$
\frac{d^n y}{d x^n}=\sin \left(\frac{n \pi}{2}+x\right)
$
View full question & answer→Question 103 Marks
Find the $n^{\text {th }}$ derivative of the following : $\log x$
AnswerLet $y=\log x$
Differentiate w.r.t. $x$
$
\frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x}
$
Differentiate $w . r . t . x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{1}{x}\right) \\
& \frac{d^2 y}{d x^2}=\frac{-1}{x^2}=\frac{(-1)^1}{x^2}
\end{aligned}
$
Differentiate $w . r . t . x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=(-1)^1 \frac{d}{d x}\left(\frac{1}{x^2}\right) \\
& \frac{d^3 y}{d x^3}=(-1)^1\left(\frac{-2}{x^3}\right)=\frac{(-1)^2 \cdot 1 \cdot 2}{x^3}
\end{aligned}
$
In general $n^{\text {th }}$ order derivative will be
$
\begin{aligned}
& \frac{d^n y}{d x^n}=\frac{(-1)^{n-1} \cdot 1 \cdot 2 \cdot 3 \ldots(n-1)}{x^n} \\
& \frac{d^n y}{d x^n}=\frac{(-1)^{n-1} \cdot(n-1) !}{x^n}
\end{aligned}
$
View full question & answer→Question 113 Marks
Find the $n^{\text {th }}$ derivative of the following : $\frac{1}{a x+b}$
AnswerLet $y=\frac{1}{a x+b}$
Differentiate w. r. t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(\frac{1}{a x+b}\right)=\frac{-1}{(a x+b)^2} \cdot \frac{d}{d x}(a x+b) \\
& \frac{d y}{d x}=\frac{(-1) \cdot a}{(a x+b)^2}
\end{aligned}
$
Differentiate w. r. t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=(-1)(a) \frac{d}{d x}\left(\frac{1}{(a x+b)^2}\right) \\
& \frac{d^2 y}{d x^2}=(-1)(a) \frac{-2}{(a x+b)^3} \cdot \frac{d}{d x}(a x+b) \\
& \frac{d^2 y}{d x^2}=\frac{(-1)^2 \cdot 2 \cdot 1 \cdot a^2}{(a x+b)^3}
\end{aligned}
$
Differentiate $w$. r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=(-1)^2 \cdot 2 \cdot 1 \cdot a^2 \cdot \frac{d}{d x}\left(\frac{1}{(a x+b)^3}\right) \\
& \frac{d^3 y}{d x^3}=(-1)^2 \cdot 2 \cdot 1 \cdot a^2 \cdot \frac{-3}{(a x+b)^4} \cdot \frac{d}{d x}(a x+b) \\
& \frac{d^3 y}{d x^3}=\frac{(-1)^3 \cdot 3 \cdot 2 \cdot 1 \cdot a^3}{(a x+b)^4}
\end{aligned}
$
In general $n^{\text {th }}$ order derivative will be
$
\begin{aligned}
& \frac{d^n y}{d x^n}=\frac{(-1)^n \cdot n \cdot(n-1) \ldots 2 \cdot 1 \cdot a^n}{(a x+b)^{n+1}} \\
& \frac{d^n y}{d x^n}=\frac{(-1)^n \cdot n ! \cdot a^n}{(a x+b)^{n+1}}
\end{aligned}
$
View full question & answer→Question 123 Marks
If $x=\sin t, y=e^{m t}$ then show that $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0$.
AnswerGiven that $x=\sin t \quad \therefore \quad t=\sin ^{-1} x$
and $y=e^{m t} \quad \therefore \quad y=e^{m \sin ^{-1} x}$
Differentiate w.r.t. $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(e^{m \sin ^{-1} x}\right)=e^{m \sin ^{-1} x} \cdot m \frac{d}{d x}\left(\sin ^{-1} x\right)$
$\frac{d y}{d x}=\frac{m \cdot e^{m \sin ^{-1} x}}{\sqrt{1-x^2}}$
$\sqrt{1-x^2} \frac{d y}{d x}=m y$
$\ldots[$ From (I)]
Squaring both sides
$
\left(1-x^2\right) \cdot\left(\frac{d y}{d x}\right)^2=m^2 y^2
$
Differentiate $w \cdot r, t . x$
$
\begin{aligned}
& \left(1-x^2\right) \frac{d}{d x}\left(\frac{d y}{d x}\right)^2+\left(\frac{d y}{d x}\right)^2 \frac{d}{d x}\left(1-x^2\right)=m^2 \frac{d}{d x}\left(y^2\right) \\
& \left(1-x^2\right) \cdot 2\left(\frac{d y}{d x}\right) \cdot \frac{d}{d x} \cdot\left(\frac{d y}{d x}\right)+\left(\frac{d y}{d x}\right)^2(-2 x)=m^2(2 y) \frac{d y}{d x}
\end{aligned}
$
$
2\left(1-x^2\right) \cdot \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}-2 x\left(\frac{d y}{d x}\right)^2=2 m^2 y \frac{d y}{d x}
$
Dividing throughout by $2 \frac{d y}{d x}$ we get,
$
\begin{gathered}
\left(1-x^2\right) \cdot \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=m^2 y \\
\therefore \quad\left(1-x^2\right) \cdot \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0
\end{gathered}
$
View full question & answer→Question 133 Marks
Find the second order derivative of the following : $e^{2 x} \sin 3 x$
AnswerLet $y=e^{2 x} \sin 3 x$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(e^{2 x} \sin 3 x\right)=e^{2 x} \frac{d}{d x}(\sin 3 x)+\sin 3 x \frac{d}{d x}\left(e^{2 x}\right) \\
& \frac{d y}{d x}=e^{2 x}(\cos 3 x)(3)+\sin 3 x\left(e^{2 x}\right)(2) \\
& \frac{d y}{d x}=e^{2 x}(3 \cos 3 x+2 \sin 3 x)
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left[e^{2 x}(3 \cos 3 x+2 \sin 3 x)\right] \\
& \frac{d^2 y}{d x^2}=e^{2 x} \frac{d}{d x}(3 \cos 3 x+2 \sin 3 x)+(3 \cos 3 x+2 \sin 3 x) \frac{d}{d x}\left(e^{2 x}\right) \\
& =e^{2 x}[3(-\sin 3 x)(3)+2(\cos 3 x)(3)]+(3 \cos 3 x+2 \sin 3 x) e^{2 x}(2) \\
& =e^{2 x}[-9 \sin 3 x+6 \cos 3 x+6 \cos 3 x+4 \sin 3 x] \\
& \frac{d^2 y}{d x^2}=e^{2 x}[12 \cos 3 x-5 \sin 3 x] \\
&
\end{aligned}
$
View full question & answer→Question 143 Marks
Find the second order derivative of the following : $x^2 e^x$
AnswerLet $y=x^2 e^x$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(x^2 e^x\right) \\
& \frac{d y}{d x}=x^2 \frac{d}{d x}\left(e^x\right)+e^x \frac{d}{d x}\left(x^2\right) \\
& \frac{d y}{d x}=x^2 e^x+2 x e^x=e^x\left(x^2+2 x\right)
\end{aligned}
$
Differentiate w.r. $t . x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left[e^x\left(x^2+2 x\right)\right] \\
& \frac{d^2 y}{d x^2}=e^x \frac{d}{d x}\left(x^2+2 x\right)+\left(x^2+2 x\right) \frac{d}{d x}\left(e^x\right) \\
& =e^x(2 x+2)+\left(x^2+2 x\right)\left(e^r\right) \\
& =\left(x^2+4 x+2\right) e^x \\
& \frac{d^2 y}{d x^2}=\left(x^2+4 x+2\right) e^x \\
\end{aligned}
$
View full question & answer→Question 153 Marks
Find the derivative of $\cos ^{-1} x$ w. r. t. $\sqrt{1-x^2}$.
AnswerLet $u=\cos ^{-1} x$ and $v=\sqrt{1-x^2}$, then we have to find $\frac{d u}{d v}$.
i.e. $\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$
Now, $u=\cos ^{-1} x$
Differentiate w.r. t. $x$
$\frac{d u}{d x}=\frac{d}{d x}\left(\cos ^{-1} x\right)=-\frac{1}{\sqrt{1-x^2}} \quad \ldots$ (II)
And, $v=\sqrt{1-x^2}$
Differentiate $w . r . t . x$
$
\begin{aligned}
\frac{d v}{d x} & =\frac{d}{d x}\left(\sqrt{1-x^2}\right)=\frac{1}{2 \sqrt{1-x^2}} \cdot \frac{d}{d x}\left(1-x^2\right)=\frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x) \\
\frac{d v}{d x}=-\frac{x}{\sqrt{1-x^2}} & \ldots \text { (III) }
\end{aligned}
$
Substituting (II) and (III) in (I) we get,
$
\frac{d u}{d v}=\frac{-\frac{1}{\sqrt{1-x^2}}}{-\frac{x}{\sqrt{1-x^2}}} \quad \therefore \quad \frac{d u}{d v}=\frac{1}{x}
$
View full question & answer→Question 163 Marks
If $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$, then show that $\frac{d y}{d x}=\frac{b^2 x}{a^2 y}$.
AnswerGiven that, $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$ i.e. $\frac{x}{a}=t-\frac{1}{t} \ldots$
(I) and $\frac{y}{b}=t+\frac{1}{t} \ldots$
Square of (I) - Square of (II) gives,
$
\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=\left(t-\frac{1}{t}\right)^2-\left(t+\frac{1}{t}\right)^2 \\
& =t^2-2+\frac{1}{t^2}-t^2-2-\frac{1}{t^2} \\
& \therefore \quad \frac{x^2}{a^2}-\frac{y^2}{a^2}=-4 \\
& \text { Differentiate w.r.t. } x \\
& \frac{1}{a^2} \cdot \frac{d}{d x}\left(x^2\right)-\frac{1}{b^2} \cdot \frac{d}{d x}\left(y^2\right)=\frac{d}{d x}(-4) \\
& \frac{1}{a^2}(2 x)-\frac{1}{b^2}(2 y) \cdot \frac{d}{d x}=0 \\
& \frac{1}{a^2}(2 x)-\frac{1}{b^2}(2 y) \cdot \frac{d y}{d x}=0 \\
& \frac{2 y}{b^2} \cdot \frac{d y}{d x}=\frac{2 x}{a^2} \Rightarrow \frac{d y}{d x}=\frac{b^2 x}{a^2 y} \\
& \therefore \quad \frac{d y}{d x}=\frac{b^2 x}{a^2 y}
\end{aligned}
$
View full question & answer→Question 173 Marks
If $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then show that $x^3 y \frac{d y}{d x}=1$.
AnswerGiven that, $x^4+y^4=t^2+\frac{1}{t^2}$
And
$
x^2+y^2=t+\frac{1}{t}
$
Squaring both sides,
$
\begin{aligned}
& \left(x^2+y^2\right)^2=\left(t+\frac{1}{t}\right)^2 \\
& x^4+2 x^2 y^2+y^4=t^2+2+\frac{1}{t^2} \\
& x^4+2 x^2 y^2+y^4=x^4+y^4+2 \\
& 2 x^2 y^2=2 \quad \therefore \quad x^2 y^2=1
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(x^2 y^2\right)=\frac{d}{d x}(1) \\
& x^2 \frac{d}{d x}\left(y^2\right)+y^2 \frac{d}{d x}\left(x^2\right)=0 \\
& x^2(2 y) \frac{d y}{d x}+y^2(2 x)=0 \\
& 2 x^2 y \frac{d y}{d x}=-2 x y^2 \Rightarrow \frac{d y}{d x}=-\frac{2 x y^2}{2 x^2 y} \\
& \qquad \frac{d y}{d x}=-\frac{x\left(-\frac{1}{x^2}\right)}{x^2 y} \ldots[\text { From (II)] } \\
& \quad \frac{d y}{d x}=\frac{1}{x^3 y} \quad \therefore \quad x^3 y \frac{d y}{d x}=1
\end{aligned}
$
View full question & answer→Question 183 Marks
Find $\frac{d y}{d x}$ if : $x=3 \cos t-2 \cos ^3 t, y=3 \sin t-2 \sin ^3 t$, at $t=\frac{\pi}{6}$
AnswerGiven, $y=3 \sin t-2 \sin ^3 t$
Differentiate w.r.t.t
$
\begin{aligned}
\frac{d y}{d t} & =\frac{d}{d t}\left(3 \sin t-2 \sin ^3 t\right) \\
& =3 \frac{d}{d t}(\sin t)-2(\sin t)^3 \\
& =3 \cos t-2(3) \sin ^2 t \frac{d}{d t}(\sin t) \\
& =3 \cos t-6 \sin ^2 t(\cos t) \\
& =3 \cos t\left(1-2 \sin ^2 t\right) \\
\frac{d y}{d t} & =3 \cos t \cos 2 t
\end{aligned}
$
And, $x=3 \cos t-2 \cos ^3 t$
Differentiate w.r.t.t
$
\begin{aligned}
\frac{d x}{d t} & =\frac{d}{d t}\left(3 \cos t-2 \cos ^3 t\right) \\
& =3 \frac{d}{d t}(\cos t)-2 \frac{d}{d t}\left(\cos ^3 t\right) \\
& =3(-\sin t)-2(3) \cos ^2 t \frac{d}{d t}(\cos t) \\
& =-3 \sin t-6 \cos ^2 t(-\sin t) \\
& =6 \cos ^2 t \sin t-3 \sin t \\
& =3 \sin t\left(2 \cos ^2 t-1\right) \\
\frac{d x}{d t} & =3 \sin t \cos 2 t
\end{aligned}
$
$
\therefore \quad \frac{d y}{d x}=-\cot t
$
At $t=\frac{\pi}{6}$, we get
$
\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{6}}=-\cot \left(\frac{\pi}{6}\right)=\sqrt{3}
$
View full question & answer→Question 193 Marks
Find $\frac{d y}{d x}$ if : $x=t+\frac{1}{t}, y=\frac{1}{t^2}$, at $t=\frac{1}{2}$
AnswerGiven, $y=\frac{1}{t^2}$
Differentiate $w . r . t . t$
$\frac{d y}{d t}=\frac{d}{d t}\left(\frac{1}{t^2}\right)$
$\frac{d y}{d t}=-\frac{2}{t^3}$
And, $x=t+\frac{1}{t}$
Differentiate $w . r . t . t$
$
\begin{aligned}
& \frac{d x}{d t}=\frac{d}{d t}\left(t+\frac{1}{t}\right)=1-\frac{1}{t^2} \\
& \frac{d x}{d t}=-\frac{t^2-1}{t^2}
\end{aligned}
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-\frac{2}{t^3}}{\frac{t^2-1}{t^2}} \quad \ldots$ [From (I) and (II)]
$
\therefore \quad \frac{d y}{d x}=-\frac{2}{t\left(t^2-1\right)}
$
At $t=\frac{1}{2}$, we get
$
\begin{aligned}
\left(\frac{d y}{d x}\right)_{t=\frac{1}{2}} & =-\frac{2}{\left(\frac{1}{2}\right)\left[\left(\frac{1}{2}\right)^2-1\right]} \\
& =-\frac{2}{\left(\frac{1}{2}\right)\left(\frac{1}{4}-1\right)} \\
\left(\frac{d y}{d x}\right)_{t=\frac{1}{2}} & =-\frac{2}{\left(\frac{1}{2}\right)\left(-\frac{3}{4}\right)} \\
\left(\frac{d y}{d x}\right)_{t=\frac{1}{2}} & =\frac{16}{3}
\end{aligned}
$
View full question & answer→Question 203 Marks
Find $\frac{d y}{d x}$ if : $x=\sec ^2 \theta, y=\tan ^3 \theta$, at $\theta=\frac{\pi}{3}$
AnswerGiven, $y=\tan ^3 \theta$
Differentiate w.r. t. $\theta$
$\frac{d y}{d \theta}=\frac{d}{d \theta}(\tan \theta)^3=3 \tan ^2 \theta \frac{d}{d \theta}(\tan \theta) \quad \therefore \quad \frac{d y}{d \theta}=3 \tan ^2 \theta \cdot \sec ^2 \theta$
And, $x=\sec ^2 \theta$
Differentiate w.r.t. $\theta$
$
\begin{aligned}
& \frac{d x}{d \theta}=\frac{d}{d \theta}\left(\sec ^2 \theta\right)=2 \sec \theta \cdot \frac{d}{d \theta}(\sec \theta) \\
& \frac{d x}{d \theta}=2 \sec \theta \cdot \sec \theta \tan \theta=2 \sec ^2 \theta \cdot \tan \theta
\end{aligned}
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 \tan ^2 \theta \cdot \sec ^2 \theta}{2 \sec ^2 \theta \cdot \tan \theta}$
$\ldots[$ From (I) and (II)]
$
\therefore \quad \frac{d y}{d x}=\frac{3}{2} \tan \theta
$
At $\theta=\frac{\pi}{3}$, we get
$
\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{3}}=\frac{3}{2} \tan \left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{2}
$
View full question & answer→Question 213 Marks
Find $\frac{d y}{d x}$ if : $x=\sqrt{1-t^2}, y=\sin ^{-1} t$
AnswerGiven, $y=\sin ^{-1} t$
Differentiate w.r.t.t
$
\begin{aligned}
& \frac{d y}{d t}=\frac{d}{d t}\left(\sin ^{-1} t\right)=\frac{1}{\sqrt{1-t^2}} \\
& \frac{d y}{d t}=\frac{1}{\sqrt{1-t^2}} \\
& \text { And, } x=\sqrt{1-t^2}
\end{aligned}
$
And, $x=\sqrt{1-t^2}$
Differentiate w.r.t.t
$
\begin{aligned}
& \frac{d x}{d t}=\frac{d}{d t}\left(\sqrt{1-t^2}\right)=\frac{1}{2 \sqrt{1-t^2}} \cdot \frac{d}{d t}\left(1-t^2\right) \\
& \left.\frac{d x}{d t}=\frac{1}{2 \sqrt{1-t^2}} \cdot(-2 t)=-\frac{t}{\sqrt{1-t^2}} \ldots \ldots(1)\right)
\end{aligned}
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{1}{\sqrt{1-t^2}}}{-\frac{t}{\sqrt{1-t^2}}} \ldots[$ From (I) and (II) $]$
$
\therefore \quad \frac{d y}{d x}=-\frac{1}{t}
$
View full question & answer→Question 223 Marks
Find $\frac{d y}{d x}$ if : $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$
AnswerGiven, $y=a(1-\cos \theta)$
Differentiate w.r. t. $\theta$
$
\begin{aligned}
& \frac{d y}{d \theta}=a \frac{d}{d \theta}[(1-\cos \theta)]=a[0-(-\sin \theta)] \\
& \frac{d y}{d t}=a \sin \theta
\end{aligned}
$
And, $x=a(\theta+\sin \theta)$
Differentiate w.r.t. $\theta$
$
\begin{aligned}
& \frac{d x}{d t}=a \frac{d}{d \theta}(\theta+\sin \theta)=a(1+\cos \theta) \\
& \frac{d x}{d t}=a(1+\cos \theta)
\end{aligned}
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a \sin \theta}{a(1+\cos \theta)} \ldots[$ From (I) and
$
\therefore \quad \frac{d y}{d x}=\frac{2 \sin \left(\frac{\theta}{2}\right) \cdot \cos \left(\frac{\theta}{2}\right)}{2 \cos ^2\left(\frac{\theta}{2}\right)}=\tan \left(\frac{\theta}{2}\right)
$
View full question & answer→Question 233 Marks
If $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}}$, then show that $\frac{d y}{d x}=\frac{\sec ^2 x}{2 y-1}$.
AnswerGiven that : $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}}$
Squaring both sides, we get
$
\begin{aligned}
& y^2=\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}, \text { which is same as } \\
& y^2=\tan x+\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}} \\
& y^2=\tan x+y \quad \ldots \ldots[\text { From (I) ] }
\end{aligned}
$
[From (I) ]
Differentiate $w . r . t . x$
$
\frac{d}{d x}\left(y^2\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x}
$
$
\begin{aligned}
& 2 y \frac{d y}{d x}-\frac{d y}{d x}=\sec ^2 x \\
& (2 y-1) \frac{d y}{d x}=\sec ^2 x \\
\therefore \quad & \frac{d y}{d x}=\frac{\sec ^2 x}{2 y-1}
\end{aligned}
$
View full question & answer→Question 243 Marks
If $\sec ^{-1}\left(\frac{x^3+y^3}{x^3-y^3}\right)=2 a$, then show that $\frac{d y}{d x}=\frac{x^2 \tan ^2 a}{y^2}$, where $a$ is a constant.
AnswerGiven that : $\sec ^{-1}\left(\frac{x^3+y^3}{x^3-y^3}\right)=2 a \quad \ldots$. [We will not eliminate $a$, as answer contains $a$ ]
$
\begin{aligned}
\therefore \quad & \cos ^{-1}\left(\frac{x^3-y^3}{x^3+y^3}\right)=2 a \\
& \frac{x^3-y^3}{x^3+y^3}=\cos 2 a \\
& x^3-y^3=x^3 \cos 2 a+y^3 \cos 2 a \\
& x^3-x^3 \cos 2 a=y^3 \cos 2 a+y^3 \\
& x^3(1-\cos 2 a)=y^3(1+\cos 2 a) \\
& y^3=\left(\frac{1-\cos 2 a}{1+\cos 2 a}\right) x^3 \\
& y^3=\left(\frac{2 \sin ^2 a}{2 \cos ^2 a}\right) x^3 \\
& y^3=\left(\tan ^2 a\right) x^3
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(y^3\right)=\left(\tan ^2 a\right) \frac{d}{d x}\left(x^3\right) \\
& 3 y^2 \frac{d y}{d x}=\left(\tan ^2 a\right) 3 x^2 \\
\therefore \quad & \frac{d y}{d x}=\frac{x^2 \tan ^2 a}{y^2}
\end{aligned}
$
View full question & answer→Question 253 Marks
If $\sin \left(\frac{p x^m-q y^m}{p x^m+q y^m}\right)=r$, then show that $\frac{d y}{d x}=\frac{y}{x}$, where $r$ is a constant.
AnswerGiven that : $\sin \left(\frac{p x^m-q y^m}{p x^m+q y^m}\right)=r$
$
\begin{aligned}
& \frac{p x^m-q y^m}{p x^m+q y^m}=\sin ^{-1} r \\
& \frac{p x^m-q y^m}{p x^m+q y^m}=t \quad \ldots \ldots\left[\text { Let } t=\sin ^{-1} r\right] \\
& p x^m-q y^m=p t x^m+q t y^m \\
& p x^m-p t x^m=q y^m+q t y^m \\
& p(1-t) x^m=q(1+t) y^m \\
& y^m=\left(\frac{p(1-t)}{q(1+t)}\right) x^m \\
& y^m=s \cdot x^m \\
& \text { Differentiate w.r.t. } x \\
& \frac{d}{d x}\left(y^m\right)=s \frac{d}{d x}\left(x^m\right) \\
& m y^{m-1} \frac{d y}{d x}=s \cdot m x^{m-1} \\
& \frac{d y}{d x}=s \cdot \frac{x^{m-1}}{y^{m-1}} x^{m-1} \\
& \frac{d y}{d x}=\frac{y^m}{x^m} \times \frac{x^{m-1}}{y^{m-1}} \quad \ldots \ldots[\text { From (I) }] \\
& \therefore \quad \frac{d y}{d x}=\frac{y}{x} \\
\end{aligned}
$
View full question & answer→Question 263 Marks
Find $\frac{d y}{d x}$ if : $x^2+e^{x y}=y^2+\log (x+y)$
AnswerGiven that : $x^2+e^{x y}=y^2+\log (x+y)$
Recall that $\frac{d}{d x} g(f(x))=g^{\prime}(f(x)) \cdot \frac{d}{d x} f(x)$
Differentiate $w . r . t . x$.
$
\begin{aligned}
& \frac{d}{d x}\left(x^2\right)+\frac{d}{d x}\left[e^{x y}\right]=\frac{d}{d x}\left(y^2\right)+\frac{d}{d x}[\log (x+y)] \\
& 2 x+e^{x y} \frac{d}{d x}(x y)=2 y \frac{d y}{d x}+\frac{1}{x+y} \frac{d}{d x}(x+y) \\
& 2 x+e^{x y}\left[x \frac{d y}{d x}+y(1)\right]=2 y \frac{d y}{d x}+\frac{1}{x+y}\left[1+\frac{d y}{d x}\right] \\
& 2 x+x e^{x y} \frac{d y}{d x}+y e^{x y}=2 y \frac{d y}{d x}+\frac{1}{x+y}+\frac{1}{x+y} \cdot \frac{d y}{d x} \\
& 2 x+y e^{x y}-\frac{1}{x+y}=2 y \frac{d y}{d x}-x e^{x y} \frac{d y}{d x}+\frac{1}{x+y} \cdot \frac{d y}{d x} \\
& 2 x+y e^{x y}-\frac{1}{x+y}=\left[2 y-x e^{x y}+\frac{1}{x+y}\right] \frac{d y}{d x} \\
& \frac{2 x(x+y)+y e^{x y}(x+y)-1}{x+y}=\left[\frac{2 y(x+y)-x e^{x y}(x+y)+1}{x+y}\right] \frac{d y}{d x} \\
\therefore & \frac{d y}{d x}=\frac{2 x(x+y)+y e^{y y}(x+y)-1}{2 y(x+y)-x e^{y y}(x+y)+1}
\end{aligned}
$
View full question & answer→Question 273 Marks
Find $\frac{d y}{d x}$ if : $y^3+\cos (x y)=x^2-\sin (x+y)$
AnswerGiven that: $y^3+\cos (x y)=x^2-\sin (x+y)$
Differentiate $w . r . t . x$.
$
\begin{aligned}
& \frac{d}{d x}\left(y^3\right)+\frac{d}{d x}[\cos (x y)]=\frac{d}{d x}\left(x^2\right)-\frac{d}{d x}[\sin (x+y)] \\
& 3 y^2 \frac{d}{d x}(y)-\sin (x y) \frac{d}{d x}(x y)=2 x-\cos (x+y) \frac{d}{d x}(x+y) \\
& 3 y^2 \frac{d y}{d x}-\sin (x y)\left[x \frac{d y}{d x}+y(1)\right]=2 x-\cos (x+y)\left[1+\frac{d y}{d x}\right] \\
& 3 y^2 \frac{d y}{d x}-x \sin (x y) \frac{d y}{d x}-y \sin (x y)=2 x-\cos (x+y)-\cos (x+y) \frac{d y}{d x} \\
& 3 y^2 \frac{d y}{d x}-x \sin (x y) \frac{d y}{d x}+\cos (x+y) \frac{d y}{d x}=2 x+y \sin (x y)-\cos (x+y) \\
& {\left[3 y^2-x \sin (x y)+\cos (x+y)\right] \frac{d y}{d x}=2 x+y \sin (x y)-\cos (x+y) } \\
\therefore & \frac{d y}{d x}=\frac{2 x+y \sin (x y)-\cos (x+y)}{3 y^2-x \sin (x y)+\cos (x+y)}
\end{aligned}
$
View full question & answer→Question 283 Marks
Find $\frac{d y}{d x}$ if : $x^5+x y^3+x^2 y+y^4=4$
AnswerGiven that: $x^5+x y^3+x^2 y+y^4=4$
Differentiate $w, r, t, x$.
$
\begin{aligned}
& \frac{d}{d x}\left(x^5\right)+\frac{d}{d x}\left(x y^3\right)+\frac{d}{d x}\left(x^2 y\right)+\frac{d}{d x}\left(y^4\right)=\frac{d}{d x}(4) \\
& 5 x^4+x \frac{d}{d x}\left(y^3\right)+y^3 \frac{d}{d x}(x)+x^2 \frac{d}{d x}(y)+y \frac{d}{d x}\left(x^2\right)+4 y^3 \frac{d}{d x}(y)=0 \\
& 5 x^4+x\left(3 y^2\right) \frac{d y}{d x}+y^3(1)+x^2 \frac{d y}{d x}+y(2 x)+4 y^3 \frac{d y}{d x}=0 \\
& x^2 \frac{d y}{d x}+3 x y^2 \frac{d y}{d x}+4 y^3 \frac{d y}{d x}=-5 x^4-2 x y-y^3 \\
& \left(x^2+3 x y^2+4 y^3\right) \frac{d y}{d x}=-\left(5 x^4+2 x y+y^3\right) \\
\therefore \quad & \frac{d y}{d x}=-\frac{5 x^4+2 x y+y^3}{x^2+3 x y^2+4 y^3}
\end{aligned}
$
View full question & answer→Question 293 Marks
Differentiate the following w. r. t. x.$\left(\frac{\left(x^2+3\right)^2 \sqrt[3]{\left(x^3+5\right)^2}}{\sqrt{\left(2 x^2+1\right)^3}}\right)$
AnswerLet $\begin{aligned} y & =\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1+2-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+x-2\right)}\right) \\ & =\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+4 x-3 x-2\right)}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-[2 x(3 x+2)-(3 x+2)]}\right) \\ & =\tan ^{-1}\left(\frac{5 x+1}{1-(3 x+2)(2 x-1)}\right)=\tan ^{-1}\left(\frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)}\right)\end{aligned}$$
y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)
$
Differentiate $w . r . t . x$.
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)\right] \\
& =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)\right]+\frac{d}{d x}\left[\tan ^{-1}(2 x-1)\right] \\
& =\frac{1}{1+(3 x+2)^2} \cdot \frac{d}{d x}(3 x+2)+\frac{1}{1+(2 x-1)^2} \cdot \frac{d}{d x}(2 x-1) \\
\therefore \quad \frac{d y}{d x} & =\frac{3}{1+(3 x+2)^2}+\frac{2}{1+(2 x-1)^2}
\end{aligned}
$
View full question & answer→Question 303 Marks
Differentiate the following w. r. t. x.$\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$
Answer$
\text { (iv) } \begin{aligned}
\text { Let } y & =\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1+2-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+x-2\right)}\right) \\
& =\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+4 x-3 x-2\right)}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-[2 x(3 x+2)-(3 x+2)]}\right) \\
& =\tan ^{-1}\left(\frac{5 x+1}{1-(3 x+2)(2 x-1)}\right)=\tan ^{-1}\left(\frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)}\right)
\end{aligned}
$
$
\begin{aligned}
& y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1) \\
& \text { Differentiate w.r.t. } x \text {. } \\
& \frac{d y}{d x}=\frac{d}{d x}\left[\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)\right] \\
& =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)\right]+\frac{d}{d x}\left[\tan ^{-1}(2 x-1)\right] \\
& =\frac{1}{1+(3 x+2)^2} \cdot \frac{d}{d x}(3 x+2)+\frac{1}{1+(2 x-1)^2} \cdot \frac{d}{d x}(2 x-1) \\
& \therefore \quad \frac{d y}{d x}=\frac{3}{1+(3 x+2)^2}+\frac{2}{1+(2 x-1)^2} \\
\end{aligned}
$
View full question & answer→Question 313 Marks
Differentiate the following w. r. t. x.$\cot ^{-1}\left(\frac{b \sin x-a \cos x}{a \sin x+b \cos x}\right)$
View full question & answer→Question 323 Marks
Differentiate the following w. r. t. x.$\tan ^{-1}\left(\frac{7 x}{1-12 x^2}\right)$
View full question & answer→Question 333 Marks
Differentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{5 \sqrt{1-x^2}-12 x}{13}\right)$
View full question & answer→Question 343 Marks
Differentiate the following w. r. t. x.$\tan ^{-1}\left(\sqrt{\frac{3-x}{3+x}}\right)$
View full question & answer→Question 353 Marks
Differentiate the following w. r. t. x.$\cot ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
AnswerLet $y=\cot ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)$
$=\tan ^{-1}\left(\frac{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]^2}{\cos ^2\left(\frac{x}{2}\right)-\sin ^2\left(\frac{x}{2}\right)}\right)=\tan ^{-1}\left(\frac{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]^2}{\left[\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]}\right)$
$=\tan ^{-1}\left(\frac{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}\right)=\tan ^{-1}\left(\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right) \ldots$. Divide Numerator \& Denominator by $\cos \left(\frac{x}{2}\right)$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]$
$\therefore y=\frac{\pi}{4}+\frac{x}{2}$
Differentiate $w . r . t . x$.
$
\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}+\frac{x}{2}\right)=0+\frac{1}{2} \quad \therefore \frac{d y}{d x}=\frac{1}{2}
$
View full question & answer→Question 363 Marks
Find the nth derivative of the following:
log(ax + b)
Question 373 Marks
If $y = x +\tan x$, show that $\cos ^2 x \cdot \frac{d^2 y}{d x^2}-2 y+2 x=0$
View full question & answer→Question 383 Marks
Differentiate $\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$ w.r.t. $\sec ^{-1} x$.
View full question & answer→Question 393 Marks
Differentiate 3x w.r.t. $\log_{x_ 3. }$
View full question & answer→Question 403 Marks
Differentiate x sin x w.r.t tan x.
Question 413 Marks
If $x =\log \left(1+ t ^2\right), y = t -\tan ^{-1} t$, show that $\frac{d y}{d x}=\frac{\sqrt{e^x-1}}{2}$
View full question & answer→Question 423 Marks
If $x=2 \cos ^4(t+3), y=3 \sin ^4(t+3)$, show that $\frac{d y}{d x}=-\sqrt{\frac{3 y}{2 x}}$
View full question & answer→Question 433 Marks
If $x = a \cos ^3 t , y = a \sin ^3 t$, then show that $\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$
View full question & answer→Question 443 Marks
If $x =\frac{t+1}{t-1}, y =\frac{1-t}{t+1}$, then show that $y 2-\frac{d y}{d x}=0$.
View full question & answer→Question 453 Marks
If $x =e^{ sin 3 t}, y =e^{\cos 3 t}$, then show that $\frac{d y}{d x}=-\frac{y \log x}{x \log y}$
View full question & answer→Question 463 Marks
Find $\frac{d y}{d x}$ if$x=\cos ^{-1}\left(4 t^3-3 t\right), y=\tan ^{-1}\left(\frac{\sqrt{1-t^2}}{t}\right)$
View full question & answer→Question 473 Marks
Find $\frac{d y}{d x}$ if$x=\cos ^{-1}\left(\frac{2 t}{1+t^2}\right), y=\sec ^{-1}\left(\sqrt{1+t^2}\right)$
View full question & answer→Question 483 Marks
Find $\frac{d y}{d x}$ if$x=\left(t+\frac{1}{t}\right)^a, y=a^{t+\frac{1}{t}}$, where $a>0, a \neq 1$ and $t \neq 0$
View full question & answer→Question 493 Marks
Find $\frac{d y}{d x}$ if$x=\sqrt{a^2+m^2}, y=\log \left(a^2+m^2\right)$
View full question & answer→Question 503 Marks
If $y=x^{x^{x^{-\infty}}}$, then show that $\frac{d y}{d x}=\frac{y^2}{x(1-\log y)}$
View full question & answer→Question 513 Marks
If $x^y= e ^{x-y}$, then show that $\frac{d y}{d x}=\frac{\log x}{(1+\log x)^2}$
View full question & answer→Question 523 Marks
If $\log _5\left(\frac{x^4+y^4}{x^4-y^4}\right)=2$, show that $\frac{d y}{d x}=-\frac{12 x^3}{13 y^3}$
View full question & answer→Question 533 Marks
If $\log (x+y)=\log (x y)+p$, where $p$ is a constant, then prove that $\frac{d y}{d x}=-\frac{y^2}{x^2}$.
View full question & answer→Question 543 Marks
Find $\frac{d y}{d x}$ if$e^{e^{x-y}}=\frac{x}{y}$
View full question & answer→Question 553 Marks
Find $\frac{d y}{d x}$ if
cos (xy) = x + y
View full question & answer→Question 563 Marks
Find $\frac{d y}{d x}$ if$e^{x+y}=\cos (x-y)$
View full question & answer→Question 573 Marks
Find $\frac{d y}{d x}$ if$x e^y+y e^x=1$
View full question & answer→Question 583 Marks
Find $\frac{d y}{d x}$ if$x^2 y^2-\tan ^{-1}\left(\sqrt{x^2+y^2}\right)=\cot ^{-1}\left(\sqrt{x^2+y^2}\right)$
View full question & answer→Question 593 Marks
Find $\frac{d y}{d x}$ if$x^3+x^2 y+x y^2+y^3=81$
View full question & answer→Question 603 Marks
Find $\frac{d y}{d x}$ if$x+\sqrt{x y}+y=1$
View full question & answer→Question 613 Marks
Differentiate the following w.r.t. x: $\sin x^x$
View full question & answer→Question 623 Marks
Differentiate the following w.r.t. x: (sin x)x
Question 633 Marks
Differentiate the following w.r.t. x: $x^{\tan ^{-1} x}$
View full question & answer→Question 643 Marks
Diffrentiate the following w. r. t. x. $\cot ^{-1}\left(\frac{4-x-2 x^2}{3 x+2}\right)$
View full question & answer→Question 653 Marks
Diffrentiate the following w. r. t. x. $\tan ^{-1}\left(\frac{5-x}{6 x^2-5 x-3}\right)$
View full question & answer→Question 663 Marks
Diffrentiate the following w. r. t. x. $\tan ^{-1}\left(\frac{a+b \tan x}{b-a \tan x}\right)$
View full question & answer→Question 673 Marks
Diffrentiate the following w. r. t. x. $\cot ^{-1}\left(\frac{a^2-6 x^2}{5 a x}\right)$
View full question & answer→Question 683 Marks
Diffrentiate the following w. r. t. x. $\tan ^{-1}\left(\frac{2^x}{1+2^{2 x+1}}\right)$
View full question & answer→Question 693 Marks
Diffrentiate the following w. r. t. x. $\tan ^{-1}\left(\frac{2^{x+2}}{1-3\left(4^x\right)}\right)$
View full question & answer→Question 703 Marks
Diffrentiate the following w. r. t. x. $\tan ^{-1}\left(\frac{2 \sqrt{x}}{1+3 x}\right)$
View full question & answer→Question 713 Marks
Diffrentiate the following w. r. t. x. $\cot ^{-1}\left(\frac{1+35 x^2}{2 x}\right)$
View full question & answer→Question 723 Marks
Diffrentiate the following w. r. t. x.$\operatorname{cosec}^{-1}\left(\frac{10}{6 \sin \left(2^x\right)-8 \cos \left(2^x\right)}\right)$
View full question & answer→Question 733 Marks
Diffrentiate the following w. r. t. x.$\cos ^{-1}\left(\frac{3 \cos \left(e^x\right)+2 \sin \left(e^x\right)}{\sqrt{13}}\right)$
View full question & answer→Question 743 Marks
Diffrentiate the following w. r. t. x.$\cos ^{-1}\left(\frac{3 \cos 3 x-4 \sin 3 x}{5}\right)$
View full question & answer→Question 753 Marks
Diffrentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{\cos \sqrt{x}+\sin \sqrt{x}}{\sqrt{2}}\right)$
Answer$y=\sin ^{-1}\left(\frac{\cos \sqrt{x}+\sin \sqrt{x}}{\sqrt{2}}\right)$
$=\sin ^{-1}\left(\frac{1}{\sqrt{2}} \cos \sqrt{x}+\frac{1}{\sqrt{2}} \sin \sqrt{x}\right)$
Put,
$\frac{1}{\sqrt{2}}=\sin x$
$\frac{1}{\sqrt{2}}=\cos \alpha$
Also,
$\sin ^2 \alpha+\cos ^2 \alpha=\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2=1$
And,
$\tan \alpha=1$
$\therefore \alpha=\tan ^{-1} 1$
$y=\sin ^{-1}(\sin \alpha \cdot \cos \sqrt{x}+\cos \alpha \cdot \sin (x)$
$=\sin ^{-1}(\sin (\alpha+\sqrt{x}))$
$y=\alpha+\sqrt{x}$
$y=\tan ^{-1}(1)+\sqrt{x}$
Differentiating w.r.t. $x$, we get
$\frac{ dy }{ dx }=\frac{ d }{ dx }\left(\tan ^{-1}+\sqrt{x}\right)$
$=0+\frac{1}{2 \sqrt{x}}$
$=\frac{1}{2 \sqrt{x}} .$
View full question & answer→Question 763 Marks
Diffrentiate the following w. r. t. x.$\cos ^{-1}\left(\frac{\sqrt{3} \cos x-\sin x}{2}\right)$
AnswerLet $y=\cos ^{-1}\left(\frac{\sqrt{3} \cos x-\sin x}{2}\right)$
$=\cos ^{-1}\left[(\cos x)\left(\frac{\sqrt{3}}{2}\right)-(\sin x)\left(\frac{1}{2}\right)\right]$
$=\cos ^{-1}\left(\cos x \cos \frac{\pi}{6}-\sin x \sin \frac{\pi}{6}\right)$
$\cdots\left[\because \cos \frac{\pi}{6}=\frac{\sqrt{3}}{2}, \sin \frac{\pi}{6}=\frac{1}{2}\right]$
$=\cos ^{-1}\left[\cos \left(x+\frac{\pi}{6}\right)\right]$
$=x+\frac{\pi}{6}$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x} =\frac{d}{d x}\left(x+\frac{\pi}{6}\right)$
$=\frac{d}{d x}(x)+\frac{d}{d x}\left(\frac{\pi}{6}\right)$
$ =1+0=1 .$
View full question & answer→Question 773 Marks
Diffrentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{4 \sin x+5 \cos x}{\sqrt{41}}\right)$
AnswerLet $y=\sin ^{-1}\left(\frac{4 \sin x+5 \cos x}{\sqrt{41}}\right)$$=\sin ^{-1}\left[(\sin x)\left(\frac{4}{\sqrt{41}}\right)+(\cos x)\left(\frac{5}{\sqrt{41}}\right)\right]$
Since, $\left(\frac{4}{\sqrt{41}}\right)^2+\left(\frac{5}{\sqrt{41}}\right)^2=\frac{16}{41}+\frac{25}{41}=1$,
we can write, $\frac{4}{\sqrt{41}}=\cos \alpha$ and $\frac{5}{\sqrt{41}}=\sin \alpha$.
$ \therefore y & =\sin ^{-1}(\sin x \cos \alpha+\cos x \sin \alpha)$
$=\sin ^{-1}[\sin (x+\alpha)]$
$=x+\alpha, \quad \text { where } \alpha \text { is a constant }$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}(x+\alpha)$
$=\frac{d}{d x}(x)+\frac{d}{d x}(\alpha)$
$=1+0=1$
View full question & answer→Question 783 Marks
Diffrentiate the following w. r. t. x.
$\tan ^{-1}(\operatorname{cosec} x+\cot x)$
AnswerLet $y=\tan ^{-1}(\operatorname{cosec} x+\cot x)$
$=\tan ^{-1}\left(\frac{1}{\sin x}+\frac{\cos x}{\sin x}\right)$
$=\tan ^{-1}\left(\frac{1+\cos x}{\sin x}\right)$
$=\tan ^{-1}\left[\frac{2 \cos ^2\left(\frac{x}{2}\right)}{2 \sin \left(\frac{x}{2}\right) \cdot \cos \left(\frac{x}{2}\right)}\right]$
$=\tan ^{-1}\left[\cot \left(\frac{x}{2}\right)\right]$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right]$
$=\frac{\pi}{2}-\frac{x}{2}$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x} =\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{2}\right) $
$ =\frac{d}{d x}\left(\frac{\pi}{2}\right)-\frac{1}{2} \frac{d}{d x}(x) $
$ =0-\frac{1}{2} \times 1=-\frac{1}{2}$
View full question & answer→Question 793 Marks
Diffrentiate the following w. r. t. x.
$\tan ^{-1}\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)$
Answer$\text { Let } y =\tan ^{-1}\left(\sqrt{\frac{1+\cos x}{1-\cos x}}\right)$
$\quad=\tan ^{-1}\left(\sqrt{\frac{2 \cos ^2\left(\frac{x}{2}\right)}{2 \sin ^2\left(\frac{x}{2}\right)}}\right)$
$=\tan ^{-1}\left[\cot \left(\frac{x}{2}\right)\right]$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right]$
$=\frac{\pi}{2}-\frac{x}{2}$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{2}\right)$
$=\frac{d}{d x}\left(\frac{\pi}{2}\right)-\frac{1}{2} \frac{d}{d x}(x)$
$=0-\frac{1}{2} \times 1=-\frac{1}{2} .$
View full question & answer→Question 803 Marks
Diffrentiate the following w. r. t. x.$\tan ^{-1}\left(\frac{\cos 7 x}{1+\sin 7 x}\right)$
AnswerLet $y=\tan ^{-1}\left(\frac{\cos 7 x}{1+\sin 7 x}\right)$
$=\tan ^{-1}\left[\frac{\sin \left(\frac{\pi}{2}-7 x\right)}{1+\cos \left(\frac{\pi}{2}-7 x\right)}\right]$
$=\tan ^{-1}\left[\frac{2 \sin \left(\frac{\pi}{4}-\frac{7 x}{2}\right) \cdot \cos \left(\frac{\pi}{4}-\frac{7 x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{7 x}{2}\right)}\right]$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{7 x}{2}\right)\right]$
$=\frac{\pi}{4}-\frac{7 x}{2}$
Differentiating w.r.t. $x$, we get
$ \frac{d y}{d x} =\frac{d}{d x}\left(\frac{\pi}{4}-\frac{7 x}{2}\right)$
$ =\frac{d}{d x}\left(\frac{\pi}{4}\right)-\frac{7}{2} \frac{d}{d x}(x)$
$=0-\frac{7}{2} \times 1=-\frac{7}{2} .$
View full question & answer→Question 813 Marks
Diffrentiate the following w. r. t. x.
$\cot ^{-1}\left(\frac{\sin 3 x}{1+\cos 3 x}\right)$
AnswerLet $y =\cot ^{-1}\left(\frac{\sin 3 x}{1+\cos 3 x}\right)$
$=\cot ^{-1}\left[\frac{2 \sin \left(\frac{3 x}{2}\right) \cos \left(\frac{3 x}{2}\right)}{2 \cos ^2\left(\frac{3 x}{2}\right)}\right]$
$=\cot ^{-1}\left[\tan \left(\frac{3 x}{2}\right)\right]$
$=\cot ^{-1}\left[\cot \left(\frac{\pi}{2}-\frac{3 x}{2}\right)\right]$
$=\frac{\pi}{2}-\frac{3 x}{2}$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{3 x}{2}\right)$
$=\frac{d}{d x}\left(\frac{\pi}{2}\right)-\frac{3}{2} \frac{d}{d x}(x)$
$\quad=0-\frac{3}{2} \times 1=-\frac{3}{2}$
View full question & answer→Question 823 Marks
Diffrentiate the following w. r. t. x.
$\tan ^{-1}\left(\frac{1+\cos \left(\frac{x}{3}\right)}{\sin \left(\frac{x}{3}\right)}\right)$
AnswerLet $y=\tan ^{-1}\left[\frac{1+\cos \left(\frac{x}{3}\right)}{\sin \left(\frac{x}{3}\right)}\right]$
$=\tan ^{-1}\left[\frac{2 \cos ^2\left(\frac{x}{6}\right)}{2 \sin \left(\frac{x}{6}\right) \cos \left(\frac{x}{6}\right)}\right]$
$=\tan ^{-1}\left[\cot \left(\frac{x}{6}\right)\right]$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{6}\right)\right]$
$=\frac{\pi}{2}-\frac{x}{6}$
Differentiating w.r.t. $x$, we get
$ \frac{d y}{d x} =\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x}{6}\right)$
$ =\frac{d}{d x}\left(\frac{\pi}{2}\right)-\frac{1}{6} \frac{d}{d x}(x) $
$ =0-\frac{1}{6} \times 1=-\frac{1}{6}$
View full question & answer→Question 833 Marks
Diffrentiate the following w. r. t. x.$\operatorname{cosec}^{-1}\left(\frac{1}{4 \cos ^3 2 x-3 \cos 2 x}\right)$
AnswerLet $y=\operatorname{cosec}^{-1}\left(\frac{1}{4 \cos ^3 2 x-3 \cos 2 x}\right)$
$=\operatorname{cosec}^{-1}\left(\frac{1}{\cos 6 x}\right) \ldots\left[\because \cos 3 x=4 \cos ^3 x-3 \cos x\right]$
$=\operatorname{cosec}^{-1}(\sec 6 x)$
$=\operatorname{cosec}^{-1}\left[\operatorname{cosec}\left(\frac{\pi}{2}-6 x\right)\right]$
$=\frac{\pi}{2}-6 x$
Differentiating w.r.t. $x$, we get
$ \frac{d y}{d x} =\frac{d}{d x}\left(\frac{\pi}{2}-6 x\right)$
$ =\frac{d}{d x}\left(\frac{\pi}{2}\right)-6 \frac{d}{d x}(x)$
$ =0-6 \times 1=-6 .$
View full question & answer→Question 843 Marks
Diffrentiate the following w. r. t. x.$\tan ^{-1}\left(\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right)$
AnswerLet $y=\tan ^{-1}\left(\frac{1-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{x}{2}\right)}\right)$
$=\tan ^{-1}\left[\frac{\tan \left(\frac{\pi}{4}\right)-\tan \left(\frac{x}{2}\right)}{1+\tan \left(\frac{\pi}{4}\right) \cdot \tan \left(\frac{x}{2}\right)}\right] \cdots\left[\because \tan \frac{\pi}{4}=1\right]$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right]$
$=\frac{\pi}{4}-\frac{x}{2}$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}-\frac{x}{2}\right)$
$=\frac{d}{d x}\left(\frac{\pi}{4}\right)-\frac{1}{2} \frac{d}{d x}(x)$
$=0-\frac{1}{2} \times 1=-\frac{1}{2} .$
View full question & answer→Question 853 Marks
Diffrentiate the following w. r. t. x.$\cos ^{-1}\left(\sqrt{\frac{1-\cos \left(x^2\right)}{2}}\right)$
AnswerLet $y=\cos ^{-1}\left(\sqrt{\frac{1-\cos \left(x^2\right)}{2}}\right)$
$=\cos ^{-1}\left(\sqrt{\frac{2 \sin ^2\left(\frac{x^2}{2}\right)}{2}}\right)$
$=\cos ^{-1}\left[\sin \left(\frac{x^2}{2}\right)\right]$
$=\cos ^{-1}\left[\cos \left(\frac{\pi}{2}-\frac{x^2}{2}\right)\right]$
$=\frac{\pi}{2}-\frac{x^2}{2}$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-\frac{x^2}{2}\right)$
$=\frac{d}{d x}\left(\frac{\pi}{2}\right)-\frac{1}{2} \frac{d}{d x}\left(x^2\right)$
$=0-\frac{1}{2} \times 2 x=-x$
View full question & answer→Question 863 Marks
Diffrentiate the following w. r. t. x.$\cos ^{-1}\left(\sqrt{\frac{1+\cos x}{2}}\right)$
AnswerLet$\begin{aligned} y & =\cos ^{-1}\left(\sqrt{\frac{1+\cos x}{2}}\right) \\ & =\cos ^{-1}\left(\sqrt{\frac{2 \cos ^2\left(\frac{x}{2}\right)}{2}}\right) \\ & =\cos ^{-1}\left[\cos \left(\frac{x}{2}\right)\right] \\ & =\frac{x}{2} \end{aligned} $Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(\frac{x}{2}\right)=\frac{1}{2} \frac{d}{d x}(x) \\ & =\frac{1}{2} \times 1=\frac{1}{2} .\end{aligned}$
View full question & answer→Question 873 Marks
Using derivative prove$\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2} \ldots[$ for $|x| \geq 1]$
AnswerLet $f(x)=\sec ^{-1} x+\operatorname{cosec}^{-1} x$ for $|x| \geq 1 \ldots . .(1)$Differentiating w.r.t. x, we get
$f^{\prime}(x)=\frac{d}{d x}\left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right)$
$=\frac{d}{d x}\left(\sec ^{-1} x\right)+\frac{d}{d x}\left(\operatorname{cosec}^{-1} x\right)$
$=\frac{1}{x \sqrt{x^2-1}}-\frac{1}{x \sqrt{x^2-1}}=0$.
Since, f'(x) = 0, f(x) is a constant function. Let f(x) = k. For any value of x, f(x) = k, where |x| > 1 Let x = 2. Then, f(2) = k ……(2)
From (1), $f(2)=\sec ^{-1}(2)+\operatorname{cosec}^{-1}(2)=\frac{\pi}{3}+\frac{\pi}{6}=\frac{\pi}{2}$
$\therefore k=\frac{\pi}{2} \quad \ldots$ [By (2)]
$\therefore f(x)=k=\frac{\pi}{2}$
Hence, $\sec ^{-1} x+\operatorname{cosec}^{-1} x=\frac{\pi}{2} . \quad \ldots$ [By (1)]
View full question & answer→Question 883 Marks
Using derivative prove $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$
Answerlet $f(x)=\tan ^{-1} x+\cot ^{-1} x$Differentiating w.r.t. x, we get
$\begin{aligned} f^{\prime}(x) & =\frac{d}{d x}\left(\tan ^{-1} x+\cot ^{-1} x\right) \\ & =\frac{d}{d x}\left(\tan ^{-1} x\right)+\frac{d}{d x}\left(\cot ^{-1} x\right) \\ & =\frac{1}{1+x^2}-\frac{1}{1+x^2}=0\end{aligned}$
Since, f'(x) = 0, f(x) is a constant function. Let f(x) = k. For any value of x, f(x) = k Let x = 0. Then f(0) = k ….(2) From (1), f(0) = tan-1(0) + cot-1(0)
$=0+\frac{\pi}{2}=\frac{\pi}{2}$
$\therefore k=\frac{\pi}{2} \quad \ldots[$ By (2)]
$\therefore f(x)=k=\frac{\pi}{2}$
Hence, $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2} . \quad \ldots$ [By (1)]
View full question & answer→Question 893 Marks
Find the derivative of the inverse of the following functions, and also fid their value at the points indicated against them :
$y=\sin (x-2)+x^2$, at $x=2$
Answer$y=\sin (x-2)+x^2$Differentiating w.r.t. x, we get
$ \frac{d y}{d x} =\frac{d}{d x}\left[\sin (x-2)+x^2\right]$
$ =\frac{d}{d x}[\sin (x-2)]+\frac{d}{d x}\left(x^2\right)$
$ =\cos (x-2) \cdot \frac{d}{d x}(x-2)+2 x$
$=\cos (x-2) \cdot(1-0)+2 x$
$ =\cos (x-2)+2 x$
The derivative of inverse function of $y=f(x)$ is
given by
$ \frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)} =\frac{1}{\cos (x-2)+2 x}$
$\text { At } x=2, \frac{d x}{d y} =\left(\frac{1}{[\cos (x-2)+2 x]}\right)_{\text {at } x=2}$
$ =\frac{1}{\cos 0+2(2)}=\frac{1}{1+4}=\frac{1}{5}$
View full question & answer→Question 903 Marks
Find the derivative of the inverse of the following functions, and also fid their value at the points indicated against them :
$y=3 x^2+2 \log x^3$, at $x=1$
Answer$y=3 x^2+2 \log x^3$
$=3 x^2+6 \log x$
Differentiating w.r.t. x, we get
$\frac{d y}{d x} =\frac{d}{d x}\left(3 x^2+6 \log x\right)$
$ =3 \frac{d}{d x}\left(x^2\right)+6 \frac{d}{d x}(\log x)$
$=\frac{6 x^2+6}{x}$
The derivative of inverse function of $y = f(x)$ is given by
$\frac{d x}{d y}=\frac{1}{\left(\frac{d y}{d x}\right)}=\frac{1}{\left(\frac{6 x^2+6}{x}\right)}$
$=\frac{x}{6 x^2+6}$
$\text { At } x=1, \frac{d x}{d y}=\left(\frac{x}{6 x^2+6}\right)_{\text {at } x=1}$
$=\frac{1}{6(1)^2+6}=\frac{1}{12} .$
View full question & answer→Question 913 Marks
If $h ( x )=\sqrt{4 f(x)+3 g(x)}, f (1)=4, g (1)=3, f ^{\prime}(1)=3, g ^{\prime}(1)=4$ find $h ^{\prime}(1)$.
AnswerGiven f(1) = 4, g(1) = 3, f ‘(1) = 3, g'(1) = 4 …..(1)
Now, $h(x)=\sqrt{4 f(x)+3 g(x)}$
$\begin{aligned} \therefore h^{\prime}(x) & =\frac{d}{d x}[\sqrt{4 f(x)+3 g(x)}] \\ & =\frac{1}{2 \sqrt{4 f(x)+3 g(x)}} \cdot \frac{d}{d x}[4 f(x)+3 g(x)] \\ & =\frac{1}{2 \sqrt{4 f(x)+3 g(x)}} \times\left[4 f^{\prime}(x)+3 g^{\prime}(x)\right] \\ \therefore h^{\prime}(1) & =\frac{1}{2 \sqrt{4 f(1)+3 g(1)}} \times\left[4 f^{\prime}(1)+3 g^{\prime}(1)\right] \\ & =\frac{1}{2 \sqrt{4 \times 4+3 \times 3}} \times[4 \times 3+3 \times 4] \ldots[\text { By }(1)] \\ & =\frac{1}{2 \sqrt{25}} \times 24 \\ & =\frac{1}{2 \times 5} \times 24=\frac{12}{5} .\end{aligned}$
View full question & answer→Question 923 Marks
Diffrentiate the following w.r.t.x$\frac{\left(x^2+2\right)^4}{\sqrt{x^2+5}}$
AnswerLet $y=\frac{\left(x^2+2\right)^4}{\sqrt{x^2+5}}$Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{\left(x^2+2\right)^4}{\sqrt{x^2+5}}\right]$
$\begin{aligned} & =\frac{\sqrt{x^2+5} \cdot \frac{d}{d x}\left(x^2+2\right)^4-\left(x^2+2\right)^4 \cdot \frac{d}{d x}\left(\sqrt{x^2+5}\right)}{\left(\sqrt{x^2+5}\right)^2} \\ & \sqrt{x^2+5} \times 4\left(x^2+2\right)^3 \cdot \frac{d}{d x}\left(x^2+2\right)- \\ & =\frac{\left(x^2+2\right)^4 \times \frac{1}{2 \sqrt{x^2+5}} \cdot \frac{d}{d x}\left(x^2+5\right)}{x^2+5} \\ & =\frac{\sqrt{x^2+5} \times 4\left(x^2+2\right)^3 \cdot(2 x+0)-\frac{\left(x^2+2\right)^4}{2 \sqrt{x^2+5}} \times(2 x+0)}{x^2+5} \\ & =\frac{8 x\left(x^2+5\right)\left(x^2+2\right)^3-x\left(x^2+2\right)^4}{\left(x^2+5\right)^{\frac{3}{2}}} \\ & =\frac{x\left(x^2+2\right)^3\left[8\left(x^2+5\right)-\left(x^2+2\right)\right]}{\left(x^2+5\right)^{\frac{3}{2}}} \\ & =\frac{x\left(x^2+2\right)^3\left(8 x^2+40-x^2-2\right)}{\left(x^2+5\right)^{\frac{3}{2}}} \\ & =\frac{x\left(x^2+2\right)^3\left(7 x^2+38\right)}{\left(x^2+5\right)^{\frac{3}{2}}} . \\ & \end{aligned}$
View full question & answer→Question 933 Marks
Diffrentiate the following w.r.t.x$y=(25)^{\log _5(\sec x)}-(16)^{\log _4(\tan x)}$
Answer$\begin{aligned} & y=(25)^{\log _5(\sec x)}-(16)^{\log _4(\tan x)} \\ & =5^{2 \log _5(\sec x)}-4^{2 \log _4(\tan x)} \\ & =5^{\log _5(\sec 5 x)}-4^{\log _4(\tan 2 x)} \\ & =\sec ^2 x-\tan ^2 x \ldots[\because=x] \\ & \therefore y=1\end{aligned}$Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}(1)=0$
View full question & answer→Question 943 Marks
Diffrentiate the following w.r.t.x$\log \left[\frac{a^{\cos x}}{\left(x^2-3\right)^3 \log x}\right]$
Answer$\begin{gathered}\quad \text { Let } y=\log \left[\frac{a^{\cos x}}{\left(x^2-3\right)^3 \log x}\right] \\ =\log a^{\cos x}-\log \left(x^2-3\right)^3-\log (\log x) \\ =(\cos x)(\log a)-3 \log \left(x^2-3\right)-\log (\log x)\end{gathered}$Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}\left[(\cos x)(\log a)-3 \log \left(x^2-3\right)-\log (\log x)\right]$
$\begin{aligned} & =(\log a) \cdot \frac{d}{d x}(\cos x)-3 \frac{d}{d x}\left[\log \left(x^2-3\right)\right]-\frac{d}{d x}[\log (\log x)] \\ & =(\log a)(-\sin x)-3 \times \frac{1}{x^2-3} \cdot \frac{d}{d x}\left(x^2-3\right)-\frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \\ & =-(\sin x)(\log a)-\frac{3}{x^2-3} \times(2 x-0)-\frac{1}{\log x} \times \frac{1}{x} \\ & =-(\sin x)(\log a)-\frac{6 x}{x^2-3}-\frac{1}{x \log x} .\end{aligned}$
View full question & answer→Question 953 Marks
Diffrentiate the following w.r.t.x$\log \left[\frac{e^{x^2}(5-4 x)^{\frac{3}{2}}}{\sqrt[3]{7-6 x}}\right]$
AnswerLet $y=\log \left[\frac{e^{x^2}(5-4 x)^{\frac{3}{2}}}{\sqrt[3]{7-6 x}}\right]$Using $ \log (A \cdot B)=\log A+\log B $
$\begin{aligned} & y=\log e^{x^2}+\log \left(\frac{(5-4 x)^{\frac{3}{2}}}{\sqrt[3]{7-6 x}}\right) \\ & =\log e^{x^2}+\log (5-4 x)^{\frac{3}{2}}-\log (\sqrt[3]{7-6 x}) \\ & =x^2 \log e+\frac{3}{2} \log (5-4 x)-\log (7-6 x)^{\frac{1}{3}} \\ & =x^2+\frac{3}{2} \log (5-4 x)-\frac{1}{3} \log (7-6 x)\end{aligned}$
Now, Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{ dy }{ dx }=\frac{ d }{ dx } x^2+\frac{3}{2} \frac{ d }{ dx } \log (5-4 x)-\frac{1}{3} \frac{ d }{ dx } \log (7-6 x) \\ & =2 x+\frac{3}{2} \frac{1}{5-4 x}(-4)-\frac{1}{3} \frac{1}{(7-6 x)} x(-6) \\ & =2 x-\frac{6}{(5-4 x)}+\frac{2}{(7-6 x)} \\ & 2 x-\frac{6}{5-4 x}+\frac{2}{7-6 x} .\end{aligned}$
View full question & answer→Question 963 Marks
Diffrentiate the following w.r.t.x$\log \left[4^{2 x}\left(\frac{x^2+5}{\sqrt{2 x^3-4}}\right)^{\frac{3}{2}}\right]$
Answer$\begin{aligned} & \quad \text { Let } y=\log \left[4^{2 x}\left(\frac{x^2+5}{\sqrt{2 x^3-4}}\right)^{\frac{3}{2}}\right] \\ & =\log 4^{2 x}+\log \left(\frac{x^2+5}{\sqrt{2 x^3-4}}\right)^{\frac{3}{2}} \\ & =2 x \log 4+\frac{3}{2} \log \left(\frac{x^2+5}{\sqrt{2 x^3-4}}\right) \\ & =2 x \log 4+\frac{3}{2}\left[\log \left(x^2+5\right)-\log \left(2 x^3-4\right)^{\frac{1}{2}}\right] \\ & =2 x \log 4+\frac{3}{2}\left[\log \left(x^2+5\right)-\frac{1}{2} \log \left(2 x^3-4\right)\right] \\ & =2 x \log 4+\frac{3}{2} \log \left(x^2+5\right)-\frac{3}{4} \log \left(2 x^3-4\right)\end{aligned}$Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[2 x \log 4+\frac{3}{2} \log \left(x^2+5\right)-\frac{3}{4} \log \left(2 x^3-4\right)\right] \\ & =(2 \log 4) \frac{d}{d x}(x)+\frac{3}{2} \frac{d}{d x}\left[\log \left(x^2+5\right)\right]-\frac{3}{4} \frac{d}{d x}\left[\log \left(2 x^3-4\right)\right] \\ & =(2 \log 4) \times 1+\frac{3}{2} \times \frac{1}{x^2+5} \cdot \frac{d}{d x}\left(x^2+5\right)- \\ & =2 \log 4+\frac{3}{2\left(x^2+5\right)} \times(2 x+0)-\frac{3}{4\left(2 x^3-4\right)} \times \frac{1}{2 x^3-4} \cdot \frac{d}{d x}\left(2 x^3-4\right) \\ & =2 \log 4+\frac{3 x}{x^2+5}-\frac{\left.9 x^2-0\right)}{2\left(2 x^3-4\right)} .\end{aligned}$
View full question & answer→Question 973 Marks
Diffrentiate the following w.r.t.x$\log \left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right)$
AnswerLet $\begin{aligned} y & =\log \left(\sqrt{\frac{1-\sin x}{1+\sin x}}\right) \\ & =\log \left(\sqrt{\frac{1-\sin x}{1+\sin x} \times \frac{1-\sin x}{1-\sin x}}\right) \\ & =\log \left(\sqrt{\frac{(1-\sin x)^2}{1-\sin ^2 x}}\right) \\ & =\log \left(\sqrt{\frac{(1-\sin x)^2}{\cos ^2 x}}\right)\end{aligned}$$=\log \left(\frac{1-\sin x}{\cos x}\right)$
$\begin{aligned} & =\log \left(\frac{1}{\cos x}-\frac{\sin x}{\cos x}\right) \\ & =\log (\sec x-\tan x)\end{aligned}$
Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\log (\sec x-\tan x)] \\ & =\frac{1}{\sec x-\tan x} \cdot \frac{d}{d x}(\sec x-\tan x) \\ & =\frac{1}{\sec x-\tan x} \times\left(\sec x \tan x-\sec ^2 x\right) \\ & =\frac{-\sec x(\sec x-\tan x)}{\sec x-\tan x} \\ & =-\sec x \\ & \end{aligned}$
View full question & answer→Question 983 Marks
Diffrentiate the following w.r.t.x$\log \left(\sqrt{\left.\frac{1+\cos \left(\frac{5 x}{2}\right)}{1-\cos \left(\frac{5 x}{2}\right)}\right)}\right.$
AnswerUsing $\log \left(\frac{a}{b}\right)=\log a-\log b$$\log a^b=b \log a$
$\begin{aligned} & y=\log \left(\sqrt{1+\cos \frac{5 x}{2}}\right)-\log \left(\sqrt{1-\cos \left(\frac{5 x}{2}\right)}\right) \\ & y=\log \left(1+\cos \left(\frac{x}{2}\right)^{\frac{1}{2}}-\log \left(1-\cos \left(\frac{5 x}{2}\right)\right)^{\frac{1}{2}}\right. \\ & y=\frac{1}{2} \log \left[1+\cos \left(\frac{5 x}{2}\right)\right]-\frac{1}{2} \log \left[\left(1-\cos \left(\frac{5 x}{2}\right)\right]\right.\end{aligned}$
Differentiating w.r.t. x
$\begin{aligned} & \frac{d y}{d x}=\frac{1}{2} \frac{1}{1+\cos \left(\frac{5 x}{2}\right)} \frac{d}{d x}\left(1+\cos \frac{5 x}{2}\right)-\frac{1}{2} \times \frac{1}{1-\cos \left(\frac{5 x}{2}\right)} \frac{d}{d x}\left(1-\cos \frac{5 x}{2}\right) \\ & =\frac{1}{2\left(1+\cos \left(\frac{5 x}{2}\right)\right)}\left(-\sin \left(\frac{5 x}{2}\right) \cdot \frac{5}{2}-\frac{1}{2\left(1-\cos \left(\frac{5 x}{2}\right)\right)}\left(\sin \left(\frac{5 x}{2}\right) \cdot \frac{5}{2}\right.\right. \\ & =\frac{-5 \sin \left(\frac{5 x}{2}\right)}{4\left(1+\cos \left(\frac{5 x}{2}\right)\right)}-\frac{5 \sin \left(\frac{5 x}{2}\right)}{4\left(1-\cos \left(\frac{5 x}{2}\right)\right)} \\ & =\frac{-5}{4} \sin \left(\frac{5 x}{2}\right)\left[\frac{1}{1+\cos ^2\left(\frac{5 x}{2}\right)}+\frac{1}{1-\cos ^2\left(\frac{5 x}{2}\right)}\right] \\ & =\frac{\frac{-5}{2} \sin \left(\frac{5 x}{2}\right)\left[1-\cos ^2\left(\frac{5 x}{2}\right)+1+\cos \frac{5 x}{2}\right]}{\left[1-\cos ^2\left(\frac{5 x}{2}\right)\right]} \\ & =\frac{-5}{4} \sin \left(\frac{5 x}{2}\right) \times \frac{2}{\sin ^2\left(\frac{5 x}{2}\right)} \ldots\left[\because 1-\cos ^2 x=\sin ^2 x\right] \\ & =\frac{-5}{4} \frac{1}{\sin \left(\frac{5 x}{2}\right)} \\ & -\frac{5}{2} \times \cos \cos \left(\frac{5 x}{2}\right)\end{aligned}$
View full question & answer→Question 993 Marks
Diffrentiate the following w.r.t.x$\log \left(\sqrt{\frac{1-\cos 3 x}{1+\cos 3 x}}\right)$
AnswerLet $\begin{aligned} y & =\log \left(\sqrt{\left.\frac{1-\cos 3 x}{1+\cos 3 x}\right)}\right. \\ & =\log \left(\sqrt{\frac{2 \sin ^2\left(\frac{3 x}{2}\right)}{2 \cos ^2\left(\frac{3 x}{2}\right)}}\right) \\ & =\log \tan \left(\frac{3 x}{2}\right)\end{aligned}$Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\log \tan \left(\frac{3 x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{3 x}{2}\right)} \times \frac{d}{d x}\left[\tan \left(\frac{3 x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{3 x}{2}\right)} \times \sec ^2\left(\frac{3 x}{2}\right) \cdot \frac{d}{d x}\left(\frac{3 x}{2}\right)\end{aligned}$
$\begin{aligned} & =\frac{\cos \left(\frac{3 x}{2}\right)}{\sin \left(\frac{3 x}{2}\right)} \times \frac{1}{\cos ^2\left(\frac{3 x}{2}\right)} \times \frac{3}{2} \times 1 \\ & =3 \times \frac{1}{2 \sin \left(\frac{3 x}{2}\right) \cos \left(\frac{3 x}{2}\right)} \\ & =3 \times \frac{1}{\sin 3 x}=3 \operatorname{cosec} 3 x .\end{aligned}$
View full question & answer→Question 1003 Marks
Diffrentiate the following w.r.t.x$\log \left[\tan ^3 x \cdot \sin ^4 x \cdot\left(x^2+7\right)^7\right]$
Answer$\begin{aligned} & \text { Let } y=\log \left[\tan ^3 x \cdot \sin ^4 x \cdot\left(x^2+7\right)^7\right] \\ & =\log \tan ^3 x+\log \sin ^4 x+\log \left(x^2+7\right)^7 \\ & =3 \log \tan x+4 \log \sin x+7 \log \left(x^2+7\right)\end{aligned}$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[3 \log \tan x+4 \log \sin x+7 \log \left(x^2+7\right)\right] \\ & =3 \frac{d}{d x}(\log \tan x)+4 \frac{d}{d x}(\log \sin x)+7 \frac{d}{d x}\left[\log \left(x^2+7\right)\right] \\ & =3 \times \frac{1}{\tan x} \cdot \frac{d}{d x}(\tan x)+4 \times \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)+ \\ & =3 \times \frac{1}{\tan x} \cdot \sec 2 x+4 \times \frac{1}{\sin x} \cdot \cos x+7 \times \frac{1}{x^2+7} \cdot(2 x+0) \\ & =3 \times \frac{\cos x}{\sin x} \times \frac{1}{\cos ^2 x}+4 \cot x+\frac{1}{x^2+7} \cdot \frac{d}{d x}\left(x^2+7\right) \\ & =\frac{6}{2 \sin x \cos x}+4 \cot x+\frac{14 x}{x^2+7} \\ & =\frac{6}{\sin 2 x}+4 \cot ^2 x+\frac{14 x}{x^2+7} \\ & =6 \operatorname{cosec} 2 x+4 \cot x+\frac{14 x}{x^2+7}\end{aligned}$
View full question & answer→Question 1013 Marks
Diffrentiate the following w.r.t.x$\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}$
Answerlet $y=\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}$Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left(\frac{e^{\sqrt{x}}+1}{e^{\sqrt{x}}-1}\right) \\ & =\frac{\left(e^{\sqrt{x}}-1\right) \frac{d}{d x}\left(e^{\sqrt{x}}+1\right)-\left(e^{\sqrt{x}}+1\right) \frac{d}{d x}\left(e^{\sqrt{x}}-1\right)}{\left(e^{\sqrt{x}}-1\right)^2} \\ & =\frac{\left(e^{\sqrt{x}}-1\right)\left[e^{\sqrt{x}} \cdot \frac{d}{d x}(\sqrt{x})+0\right]-\left(e^{\sqrt{x}}+1\right)\left[e^{\sqrt{x}} \cdot \frac{d}{d x}(\sqrt{x})-0\right]}{\left(e^{\sqrt{x}}-1\right)^2} \\ & =\frac{\left(e^{\sqrt{x}}-1\right)\left[e^{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}\right]-\left(e^{\sqrt{x}}+1\right)\left[e^{\sqrt{x}} \times \frac{1}{2 \sqrt{x}}\right]}{\left(e^{\sqrt{x}}-1\right)^2} \\ & =\frac{\frac{e^{\sqrt{x}}}{2 \sqrt{x}}\left(e^{\sqrt{x}}-1-e^{\sqrt{x}}-1\right)}{\left(e^{\sqrt{x}}-1\right)^2} \\ & =\frac{-e^{\sqrt{x}}}{\sqrt{x}\left(e^{\sqrt{x}}-1\right)^2} .\end{aligned}$
View full question & answer→Question 1023 Marks
Diffrentiate the following w.r.t.x$\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$
AnswerLet $y=\frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}=\frac{e^{2 x}-\frac{1}{e^{2 x}}}{e^{2 x}+\frac{1}{e^{2 x}}}=\frac{e^{4 x}-1}{e^{4 x}+1}$Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left(\frac{e^{4 x}-1}{e^{4 x}+1}\right) \\ & =\frac{\left(e^{4 x}+1\right) \cdot \frac{d}{d x}\left(e^{4 x}-1\right)-\left(e^{4 x}-1\right) \cdot \frac{d}{d x}\left(e^{4 x}+1\right)}{\left(e^{4 x}+1\right)^2} \\ & =\frac{\left(e^{4 x}+1\right)\left[e^{4 x} \cdot \frac{d}{d x}(4 x)-0\right]-\left(e^{4 x}-1\right)\left[e^{4 x} \cdot \frac{d}{d x}(4 x)+0\right]}{\left(e^{4 x}+1\right)^2}\end{aligned}$
$=\frac{\left(e^{4 x}+1\right) \cdot e^{4 x} \times 4-\left(e^{4 x}-1\right) \cdot e^{4 x} \times 4}{\left(e^{4 x}+1\right)^2 \text { } }$
$=\frac{4 e^{4 x}\left(e^{4 x}+1-e^{4 x}+1\right)}{\left(e^{4 x}+1\right)^2}=\frac{8 e^{4 x}}{\left(e^{4 x}+1\right)^2}$
View full question & answer→Question 1033 Marks
Diffrentiate the following w.r.t.x$\cot \left(\frac{\log x}{2}\right)-\log \left(\frac{\cot x}{2}\right)$
AnswerLet $y=\cot \left(\frac{\log x}{2}\right)-\log \left(\frac{\cot x}{2}\right)$Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\cot \left(\frac{\log x}{2}\right)-\log \left(\frac{\cot x}{2}\right)\right] \\ & =\frac{d}{d x}\left[\cot \left(\frac{\log x}{2}\right)\right]-\frac{d}{d x}\left[\log \left(\frac{\cot x}{2}\right)\right] \\ & =-\operatorname{cosec}^2\left(\frac{\log x}{2}\right) \cdot \frac{d}{d x}\left(\frac{\log x}{2}\right)-\frac{1}{\left(\frac{\cot x}{2}\right)} \cdot \frac{d}{d x}\left(\frac{\cot x}{2}\right) \\ & =-\operatorname{cosec}^2\left(\frac{\log x}{2}\right) \times \frac{1}{2} \times \frac{1}{x}-\frac{2}{\cot x} \times \frac{1}{2} \times\left(-\operatorname{cosec}^2 x\right) \\ & =-\frac{\operatorname{cosec}^2\left(\frac{\log x}{2}\right)}{2 x}+\tan x \cdot \operatorname{cosec} 2 x\end{aligned}$
View full question & answer→Question 1043 Marks
Diffrentiate the following w.r.t.x$\frac{1+\sin x^{\circ}}{1-\sin x^{\circ}}$
AnswerLet $y=\frac{1+\sin x^{\circ}}{1-\sin x^{\circ}}=\frac{1+\sin \left(\frac{\pi x}{180}\right)}{1-\sin \left(\frac{\pi x}{180}\right)} \quad \cdots\left[\because x^{\circ}=\left(\frac{\pi x}{180}\right)^c\right]$Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\frac{1+\sin \left(\frac{\pi x}{180}\right)}{1-\sin \left(\frac{\pi x}{180}\right)}\right) \\ & '=\frac{\left[1-\sin \left(\frac{\pi x}{180}\right)\right] \cdot \frac{d}{d x}\left[1+\sin \left(\frac{\pi x}{180}\right)\right]-\left[1+\sin \left(\frac{\pi x}{180}\right)\right] \cdot \frac{d}{d x}\left[1-\sin \left(\frac{\pi x}{180}\right)\right]}{\left[1-\sin \left(\frac{\pi x}{180}\right)\right]^2} \\ & =\frac{\left[1-\sin \left(\frac{\pi x}{180}\right)\right] \cdot\left[0+\cos \left(\frac{\pi x}{180}\right) \cdot \frac{d}{d x}\left(\frac{\pi x}{180}\right)\right]-\left[1+\sin \left(\frac{\pi x}{180}\right)\right] \cdot\left[0-\cos \left(\frac{\pi x}{180}\right) \cdot \frac{d}{d x}\left(\frac{\pi x}{180}\right)\right]}{\left[1-\sin \left(\frac{\pi x}{180}\right)\right]^2} \\ & =\frac{\left(1-\sin x^{\circ}\right)\left[\left(\cos x^{\circ}\right) \times \frac{\pi}{180} \times 1\right]-\left(1+\sin x^{\circ}\right)\left[\left(-\cos x^{\circ}\right) \times \frac{\pi}{180} \times 1\right]}{\left(1-\sin x^{\circ}\right)^2} \\ & =\frac{\frac{\pi}{180} \cos x^{\circ}\left(1-\sin x^{\circ}+1+\sin x^{\circ}\right)}{\left(1-\sin x^{\circ}\right)^2} \\ & =\frac{\pi \cos x^{\circ}}{90\left(1-\sin x^{\circ}\right)^2} . \\ & \end{aligned}$
View full question & answer→Question 1053 Marks
Diffrentiate the following w.r.t.x log(sec 3x+ tan 3x)
AnswerLet y = log(sec 3x+ tan 3x) Differentiating w.r.t. x, we get$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\log (\sec 3 x+\tan 3 x)] \\ & =\frac{1}{\sec 3 x+\tan 3 x} \cdot \frac{d}{d x}(\sec 3 x+\tan 3 x) \\ & =\frac{1}{\sec 3 x+\tan 3 x} \times\left[\frac{d}{d x}(\sec 3 x)+\frac{d}{d x}(\tan 3 x)\right] \\ & =\frac{1}{\sec 3 x+\tan 3 x} \times\left[\sec 3 x \tan 3 x \cdot \frac{d}{d x}(3 x)+\sec ^2 3 x \cdot \frac{d}{d x}(3 x)\right]\end{aligned}$
$=\frac{1}{\sec 3 x+\tan 3 x} \times\left[\sec 3 x \tan 3 x \times 3+\sec ^2 3 x \times 3\right]$
$=\frac{3 \sec 3 x(\tan 3 x+\sec 3 x)}{\sec 3 x+\tan 3 x}=3 \sec 3 x$
View full question & answer→Question 1063 Marks
Diffrentiate the following w.r.t.x$\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}$
AnswerLet $y =\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}$Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\sqrt{\cos x}+\sqrt{\cos \sqrt{x}}] \\ & =\frac{d}{d x}(\cos x)^{\frac{1}{2}}+\frac{d}{d x}(\cos \sqrt{x})^{\frac{1}{2}} \\ & =\frac{1}{2}(\cos x)^{-\frac{1}{2}} \cdot \frac{d}{d x}(\cos x)+\frac{1}{2}(\cos \sqrt{x})^{-\frac{1}{2}} \cdot \frac{d}{d x}(\cos \sqrt{x}) \\ & =\frac{1}{2 \sqrt{\cos x}} \cdot(-\sin x)+\frac{1}{2 \sqrt{\cos \sqrt{x}}} \times(-\sin \sqrt{x}) \cdot \frac{d}{d x}(\sqrt{x}) \\ & =\frac{-\sin x}{2 \sqrt{\cos x}}-\frac{\sin \sqrt{x}}{2 \sqrt{\cos \sqrt{x}}} \times \frac{1}{2 \sqrt{x}} \\ & =\frac{-\sin x}{2 \sqrt{\cos x}}-\frac{\sin \sqrt{x}}{4 \sqrt{x} \sqrt{\cos \sqrt{x}}} .\end{aligned}$
View full question & answer→Question 1073 Marks
Diffrentiate the following w.r.t.x$\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^3$
AnswerLet $y=\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^3$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^3\right] \\ & =\left(1+\sin ^2 x\right)^2 \cdot \frac{d}{d x}\left(1+\cos ^2 x\right)^3+\left(1+\cos ^2 x\right)^3 \frac{d}{d x}\left(1+\sin ^2 x\right)^2 \\ & =\left(1+\sin ^2 x\right)^2 \times 3\left(1+\cos ^2 x\right)^2 \cdot \frac{d}{d x}\left(1+\cos ^2 x\right)+\left(1+\cos ^2 x\right)^3 \times 2\left(1+\sin ^2 x\right) \cdot \frac{d}{d x}\left(1+\sin ^2 x\right) \\ & =3\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^2 \cdot\left[0+2 \cos x \cdot \frac{d}{d x}(\cos x)\right]+2\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^3 \cdot\left[0+2 \sin x \cdot \frac{d}{d x}(\sin x)\right]\end{aligned}$
$\begin{aligned} & =3\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^2 \cdot[2 \cos x(-\sin x)]+2\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^3 \cdot[2 \sin x-\cos x] \\ & =3\left(1+\sin ^2 x\right)^2\left(1+\cos ^2 x\right)^2(-\sin 2 x)+2\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^3(\sin 2 x) \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left[-3\left(1+\sin ^2 x\right)+2\left(1+\cos ^2 x\right)\right] \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left(-3-3 \sin ^2 x+2+2 \cos ^2 x\right) \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left[-1-3 \sin ^2 x+2\left(1-\sin ^2 x\right)\right] \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left(-1-3 \sin ^2 x+2-2 \sin ^2 x\right) \\ & =\sin 2 x\left(1+\sin ^2 x\right)\left(1+\cos ^2 x\right)^2\left(1-5 \sin ^2 x\right) .\end{aligned}$
View full question & answer→Question 1083 Marks
Diffrentiate the following w.r.t.x$\frac{\left(x^3-5\right)^5}{\left(x^3+3\right)^3}$
AnswerLet $y=\frac{\left(x^3-5\right)^5}{\left(x^3+3\right)^3}$Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\frac{\left(x^3-5\right)^5}{\left(x^3+3\right)^3}\right] \\ & =\frac{\left(x^3+3\right)^3 \cdot \frac{d}{d x}\left(x^3-5\right)^5-\left(x^3-5\right)^5 \cdot \frac{d}{d x}\left(x^3+3\right)^3}{\left[\left(x^3+3\right)^3\right]^2} \\ & =\frac{\left(x^3+3\right)^3 \times 5\left(x^3-5\right)^4 \cdot \frac{d}{d x}\left(x^3-5\right)-\left(x^3-5\right)^5 \times 3\left(x^3+3\right)^2 \cdot \frac{d}{d x}\left(x^3+3\right)}{\left(x^3+3\right)^6} \\ & =\frac{5\left(x^3+3\right)^3\left(x^3-5\right)^4 \cdot\left(3 x^2-0\right)-3\left(x^3-5\right)^5\left(x^3+3\right)^2 \cdot\left(3 x^2+0\right)}{\left(x^3+3\right)^6}\end{aligned}$
$=\frac{3 x^2\left(x^3+3\right)^2\left(x^3-5\right)^4\left[5\left(x^3+3\right)-3\left(x^3-5\right)\right]}{\left(x^3+3\right)^6}$
$\begin{aligned} & =\frac{3 x^2\left(x^3-5\right)^4\left(5 x^3+15-3 x^3+15\right)}{\left(x^3+3\right)^4} \\ & =\frac{3 x^2\left(x^3-5\right)^4\left(2 x^3+30\right)}{\left(x^3+3\right)^4} \\ & =\frac{6 x^2\left(x^3+15\right)\left(x^3-5\right)^4}{\left(x^3+3\right)^4}\end{aligned}$
View full question & answer→Question 1093 Marks
Diffrentiate the following w.r.t.x$\frac{x}{\sqrt{7-3 x}}$
AnswerLet $y =\frac{x}{\sqrt{7-3 x}}$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(\frac{x}{\sqrt{7-3 x}}\right)=\frac{\sqrt{7-3 x} \cdot \frac{d}{d x}(x)-x \frac{d}{d x}(\sqrt{7-3 x})}{(\sqrt{7-3 x})^2} \\ & =\frac{\sqrt{7-3 x} \times 1-x \times \frac{1}{2 \sqrt{7-3 x}} \cdot \frac{d}{d x}(7-3 x)}{7-3 x} \\ & =\frac{\sqrt{7-3 x}-\frac{x}{2 \sqrt{7-3 x}}(0-3 \times 1)}{7-3 x} \\ & =\frac{2(7-3 x)+3 x}{2(7-3 x)^{\frac{3}{2}}}=\frac{14-6 x+3 x}{2(7-3 x)^{\frac{3}{2}}}=\frac{14-3 x}{2(7-3 x)^{\frac{3}{2}}}\end{aligned}$
View full question & answer→Question 1103 Marks
Diffrentiate the following w.r.t.x$(1+4 x)^5\left(3+x-x^2\right)^8$
AnswerLet $y=(1+4 x)^5\left(3+x-x^2\right)^8$Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left[(1+4 x)^5\left(3+x-x^2\right)^8\right]$
$\begin{aligned} & =(1+4 x)^5 \cdot \frac{d}{d x}\left(3+x-x^2\right)^8+\left(3+x-x^2\right)^8 \cdot \frac{d}{d x}(1+4 x)^5 \\ & =(1+4 x)^5 \times 8\left(3+x-x^2\right)^7 \cdot \frac{d}{d x}\left(3+x-x^2\right)+\left(3+x-x^2\right)^8 \times 5(1+4 x)^4 \cdot \frac{d}{d x}(1+4 x) \\ = & 8(1+4 x)^5\left(3+x-x^2\right)^7 \cdot(0+1-2 x)+5(1+4 x)^4\left(3+x-x^2\right)^8 \cdot(0+4 \times 1) \\ = & 8(1-2 x)(1+4 x)^5\left(3+x-x^2\right)^7+20(1+4 x)^4\left(3+x-x^2\right)^8 .\end{aligned}$
View full question & answer→Question 1113 Marks
Diffrentiate the following w.r.t.x$\left(x^2+4 x+1\right)^3+\left(x^3-5 x-2\right)^4$
AnswerLet $y=\left(x^2+4 x+1\right)^3+\left(x^3-5 x-2\right)^4$Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\left( x ^2+4 x +1\right)^3+\left( x ^3-5 x -2\right)^4\right] \\ & =\frac{d}{d x}=\left( x ^2+4 x +1\right)^3+\frac{d}{d x}\left( x ^3-5 x -2\right)^4 \\ & =3\left( x ^2+4 x +1\right)^2 \cdot \frac{d}{d x}\left( x ^2+4 x +1\right)+4\left( x ^3-5 x -2\right)^4 \cdot \frac{d}{d x}\left( x ^3-5 x -2\right) \\ & =3\left( x ^2+4 x +1\right)^3 \cdot(2 x +4 \times 1+0)+4\left( x ^3-5 x -2\right)^3 \cdot\left(3 x ^2-5 \times 1-0\right) \\ & =6( x +2)\left( x ^2+4 x +1\right)^2+4\left(3 x ^2-5\right)\left( x ^3-5 x -2\right)^3 .\end{aligned}$
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