Question 12 Marks
Differentiate the following w. r. t. x.$y=a^{a^{\log _0(\cos x)}}$
Answer$
\begin{aligned}
& y=a^{a^{\log _a(\operatorname{soc} x)}} \\
& y=a^{\cot x} \quad\left[\because a^{\log _a f(x)}=f(x)\right]
\end{aligned}
$
Differentiate $w . r . t . x$
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left(a^{\cot x}\right) \\
& =a^{\cot x} \log a \cdot \frac{d}{d x}(\cot x) \\
& =a^{\cot x} \log a\left(-\operatorname{cosec}^2 x\right) \\
\frac{d y}{d x} & =-\operatorname{cosec}^2 x \cdot a^{\cot x} \log a
\end{aligned}
$
View full question & answer→Question 22 Marks
Differentiate the following w. r. t. x.$y=(4)^{\log _2(\sin x)}+(9)^{\log _y(\cos x)}$
Answer
$
\begin{aligned}
y & =(4)^{\log _2(\sin x)}+(9)^{\log _3(\cos x)} \\
& =\left(2^2\right)^{\log _2(\sin x)}+\left(3^2\right)^{\log _3(\cos x)} \\
& =(2)^{2 \log _2(\sin x)}+(3)^{2 \log _3(\cos x)} \\
& =(2)^{\log _2\left(\sin ^2 x\right)}+(3)^{\log _3\left(\cos ^2 x\right)}\left[\because a^{\log _a f(x)}=f(x)\right] \\
& =\sin ^2 x+\cos ^2 x
\end{aligned}
$
$
\therefore \quad y=1
$
Differentiate $w . r . t . x$
$
\frac{d y}{d x}=\frac{d}{d x}(1)=0
$
View full question & answer→Question 32 Marks
Differentiate the following w. r. t. x.$y=\left(x^3+2 x-3\right)^4(x+\cos x)^3$
Answer$ y=\left(x^3+2 x-3\right)^4(x+\cos x)^3 $ Differentiate $w . r . t . x$ $ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\left(x^3+2 x-3\right)^4(x+\cos x)^3\right] \\ & =\left(x^3+2 x-3\right)^4 \cdot \frac{d}{d x}(x+\cos x)^3+(x+\cos x)^3 \cdot \frac{d}{d x}\left(x^3+2 x-3\right)^4 \end{aligned} $$\begin{aligned} & =\left(x^3+2 x-3\right)^4 \cdot 3(x+\cos x)^2 \cdot \frac{d}{d x}(x+\cos x)+(x+\cos x)^3 \cdot 4\left(x^3+2 x-3\right)^3 \cdot \frac{d}{d x}\left(x^3+2 x-3\right) \\ & =\left(x^3+2 x-3\right)^4 \cdot 3(x+\cos x)^2(1-\sin x)+(x+\cos x)^3 \cdot 4\left(x^3+2 x-3\right)^3\left(3 x^2+2\right) \\ \therefore \quad \frac{d y}{d x} & =3\left(x^3+2 x-3\right)^4(x+\cos x)^2(1-\sin x)+4\left(3 x^2+2\right)\left(x^3+2 x-3\right)^3(x+\cos x)^3\end{aligned}$
View full question & answer→Question 42 Marks
Differentiate the following w. r. t. x.$y=\cot ^2\left(x^3\right)$
Answer$y=\cot ^2\left(x^3\right)$ Differentiate w. r.t. $x$ $ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(\cot ^2\left(x^3\right)\right) \\ & =\frac{d}{d x}\left[\cot \left(x^3\right)\right]^2 \\ & =2 \cot \left(x^3\right) \frac{d}{d x}\left[\cot \left(x^3\right)\right] \\ & =2 \cot \left(x^3\right)\left[-\operatorname{cosec}^2\left(x^3\right)\right] \frac{d}{d x}\left(x^3\right) \\ & =-2 \cot \left(x^3\right) \operatorname{cosec}^2\left(x^3\right)\left(3 x^2\right) \\ \therefore \quad \frac{d y}{d x} & =-6 x^2 \cot \left(x^3\right) \operatorname{cosec}^2\left(x^3\right) \end{aligned} $
View full question & answer→Question 52 Marks
Find the $n^{\text {th }}$ derivative of the following : $x^m$
AnswerLet $y=x^m$
Differentiate w.r.t. $x$
$
\frac{d y}{d x}=\frac{d}{d x}\left(x^m\right)=m x^{m-1}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=m \frac{d}{d x} x^{m-1} \\
& \frac{d^2 y}{d x^2}=m \cdot(m-1) x^{m-2}
\end{aligned}
$
Differentiate $w \cdot r \cdot t \cdot x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=m \cdot(m-1) \frac{d}{d x}\left(x^{m-2}\right) \\
& \frac{d^3 y}{d x^3}=m \cdot(m-1) \cdot(m-2) x^{m-3}
\end{aligned}
$
In general $n^{\text {th }}$ order derivative will be
$
\begin{aligned}
& \frac{d^n y}{d x^n}=m \cdot(m-1) \cdot(m-2) \ldots[m-(n-1)] x^{m-n} \\
& \frac{d^n y}{d x^n}=m \cdot(m-1) \cdot(m-2) \ldots[m-n+1] x^{m-n}
\end{aligned}
$
View full question & answer→Question 62 Marks
Find the second order derivative of the following : $\sin (\log x)$
AnswerLet $y=\sin (\log x)$
Differentiate w.r.t.x
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}[\sin (\log x)] \\
& \frac{d y}{d x}=\cos (\log x) \frac{d}{d x}(\log x) \\
& \frac{d y}{d x}=\frac{\cos (\log x)}{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[\frac{\cos (\log x)}{x}\right] \\
& \frac{d^2 y}{d x^2}=\frac{x \frac{d}{d x}[\cos (\log x)]-\cos (\log x) \frac{d}{d x}(x)}{x^2} \\
&=\frac{x[-\sin (\log x)] \frac{d}{d x}(\log x)-\cos (\log x)(1)}{x^2} \\
&=\frac{-\frac{x \sin (\log x)}{x}-\cos (\log x)}{x^2} \\
& \frac{d^2 y}{d x^2}=-\frac{\sin (\log x)+\cos (\log x)}{x^2}
\end{aligned}
$
View full question & answer→Question 72 Marks
Find the second order derivative of the following : $x^2 \log x$
AnswerLet $y=x^2 \log x$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(x^2 \log x\right) \\
& \frac{d y}{d x}=x^2 \frac{d}{d x}(\log x)+\log x \frac{d}{d x}\left(x^2\right) \\
& \frac{d y}{d x}=x^2 \cdot \frac{1}{x}+\log x(2 x) \\
& \frac{d y}{d x}=x(1+2 \log 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}[x(1+2 \log x)] \\
& \frac{d^2 y}{d x^2}=x \frac{d}{d x}(1+2 \log x)+(1+2 \log x) \frac{d}{d x}(x) \\
& \quad=x \cdot \frac{2}{x}+(1+2 \log x)(1) \\
& \frac{d^2 y}{d x^2}=3+2 \log x
\end{aligned}
$
View full question & answer→Question 82 Marks
Find the second order derivative of the following : $x^3+7 x^2-2 x-9$
AnswerLet $y=x^3+7 x^2-2 x-9$
Differentiate $w$.r.t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(x^3+7 x^2-2 x-9\right) \\
& \frac{d y}{d x}=3 x^2+14 x-2
\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(3 x^2+14 x-2)\right. \\
& \frac{d^2 y}{d x^2}=6 x+14
\end{aligned}
$
View full question & answer→Question 92 Marks
Find the derivative of $7^x$ w. r.t. $x^7$.
AnswerLet : $u=7^x$ and $v=x^7$, then we have to find $\frac{d u}{d v}$.
$
\therefore \frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}
$
Now, $u=7^x$
Differentiate w.r.t. $x$
$
\frac{d u}{d x}=\frac{d}{d x}\left(7^x\right)=7^x \log 7 \ldots
$
And, $v=x^7$
Differentiate w.r. t. $x$
$
\frac{d v}{d x}=\frac{d}{d x}\left(x^7\right)=7 x^6
$
Substituting (II) and (III) in (I) we get,
$
\therefore \frac{d u}{d v}=\frac{7^x \log 7}{7 x^6}
$
View full question & answer→Question 102 Marks
Find $\frac{d y}{d x}$ if : $x=\cos (\log t), y=\log (\cos t)$
AnswerGiven, $y=\log (\cos t)$
Differentiate $w . r . t . t$
$
\frac{d y}{d t}=\frac{d}{d t}[\log (\cos t)]=\frac{1}{\cot t} \cdot \frac{d}{d t}(\cos t)=\frac{1}{\cot t}(-\sin t) \quad \therefore \frac{d y}{d t}=-\tan t
$
And, $x=\cos (\log t)$
Differentiate w.r.t.t
$
\frac{d x}{d t}=\frac{d}{d t}[\cos (\log t)]=-\sin (\log t) \cdot \frac{d}{d t}(\log t)=-\frac{\sin (\log t)}{t} \quad \therefore \frac{d x}{d t}=-\frac{\sin (\log t)}{t} .
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-\tan t}{-\frac{\sin (\log t)}{t}} \ldots[$ From (I) and (II) ]
$
\therefore \quad \frac{d y}{d x}=\frac{t \cdot \tan t^t}{\sin (\log t)}
$
View full question & answer→Question 112 Marks
Find $\frac{d y}{d x}$ if : $x=t-\sqrt{t}, y=t+\sqrt{t}$
AnswerGiven, $y=t+\sqrt{t}$
Differentiate $w$. r.t.t
$\frac{d y}{d t}=\frac{d}{d t}(t+\sqrt{t})=1+\frac{1}{2 \sqrt{t}}$
$\frac{d y}{d t}=\frac{2 \sqrt{t}+1}{2 \sqrt{t}}$
And, $x=t-\sqrt{t}$
Differentiate $w$. r. t. $t$
$\frac{d x}{d t}=\frac{d}{d t}(t-\sqrt{t})=1-\frac{1}{2 \sqrt{t}}$
$\frac{d x}{d t}=\frac{2 \sqrt{t}-1}{2 \sqrt{t}}$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{2 \sqrt{t}+1}{2 \sqrt{t}}}{\frac{2 \sqrt{t}-1}{2 \sqrt{t}}} \ldots[$ From (I) and (II) $]$
$
\therefore \quad \frac{d y}{d x}=\frac{2 \sqrt{t}+1}{2 \sqrt{t}-1}
$
View full question & answer→Question 122 Marks
Find $\frac{d y}{d x}$ if : $x=a t^4, y=2 a t^2$
AnswerGiven, $y=2 a t^2$
Differentiate w.r.t.t
$
\frac{d y}{d t}=2 a \frac{d}{d t}\left(t^2\right)=2 a(2 t)=4 a t . \ldots
$
And, $x=a t^4$
Differentiate w.r.t.t
$
\frac{d x}{d t}=a \frac{d}{d t}\left(t^4\right)=a\left(4 t^3\right)=4 a t^3 \ldots .
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{4 a t}{4 a t^3} \ldots[$ From (I) and (II) $]$
$
\therefore \quad \frac{d y}{d x}=\frac{1}{t^2}
$
View full question & answer→Question 132 Marks
Differentiate the following w. r. t. x.$\tan ^{-1}\left(\frac{4 x}{1+21 x^2}\right)$
View full question & answer→Question 142 Marks
Differentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$
View full question & answer→Question 152 Marks
Differentiate the following w. r. t. x.$\cos ^{-1}\left(\frac{2^x-2^{-x}}{2^x+2^{-x}}\right)$
View full question & answer→Question 162 Marks
Differentiate the following w. r. t. x.$\cos ^{-1}\left(\frac{1-9 x^2}{1+9 x^2}\right)$
View full question & answer→Question 172 Marks
Differentiate the following w. r. t. x.$\operatorname{cosec}^{-1}\left(\frac{1}{3 x-4 x^3}\right)$
View full question & answer→Question 182 Marks
Differentiate the following w. r. t. x.$\cos ^{-1}\left(2 x \sqrt{1-x^2}\right)$
View full question & answer→Question 192 Marks
Differentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$
Answer$\begin{array}{ll}
\text { Let } y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right) \\
& \text { Put } x=\tan \theta \therefore \theta=\tan ^{-1} x \\
\therefore \quad & y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right) \\
& y=\sin ^{-1}(\sin 2 \theta)=2 \theta \\
\therefore \quad & y=2 \tan ^{-1} x
\end{array}
$
Differentiate $w . r . t . x$.
$
\begin{aligned}
\frac{d y}{d x} & =2 \frac{d}{d x}\left(\tan ^{-1} x\right) \\
\therefore \quad \frac{d y}{d x} & =\frac{2}{1+x^2}
\end{aligned}
$
View full question & answer→Question 202 Marks
Differentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{a \cos x-b \sin x}{\sqrt{a^2+b^2}}\right)$
Answer$\begin{aligned} & \text { Let } y=\sin ^{-1}\left(\frac{a \cos x-b \sin x}{\sqrt{a^2+b^2}}\right)=\sin ^{-1}\left(\frac{a}{\sqrt{a^2+b^2}} \cos x-\frac{b}{\sqrt{a^2+b^2}} \sin x\right) \\ & \text { Put } \frac{a}{\sqrt{a^2+b^2}}=\sin \alpha, \frac{b}{\sqrt{a^2+b^2}}=\cos \alpha \\ & \text { Also, } \sin ^2 \alpha+\cos ^2 \alpha=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=1 \text { And } \tan \alpha=\frac{a}{b} \therefore \alpha=\tan ^{-1}\left(\frac{a}{b}\right) \\ & y=\sin ^{-1}(\sin \alpha \cos x-\cos \alpha \sin x) \\ & \text { But } \sin (\alpha-x)=\sin \alpha \cos x-\cos \alpha \sin x \\ & y=\sin ^{-1}[\sin (\alpha-x)] \\ & \therefore \quad y=\tan ^{-1}\left(\frac{a}{b}\right)-x \\ & \text { Differentiate } w . r . t . x \text {. } \\ & \frac{d y}{d x}=\frac{d}{d x}\left[\tan ^{-1}\left(\frac{a}{b}\right)-x\right]=0-1 \quad \therefore \quad \frac{d y}{d x}=-1 \\ & \end{aligned}$
View full question & answer→Question 212 Marks
Differentiate the following w. r. t. x.$\cos ^{-1}\left(\frac{3 \sin x^2+4 \cos x^2}{5}\right)$
View full question & answer→Question 222 Marks
Differentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{2 \cos x+3 \sin x}{\sqrt{13}}\right)$
View full question & answer→Question 232 Marks
Differentiate the following w. r. t. x.$\cos ^{-1}\left(\sqrt{\frac{1+x}{2}}\right)$
AnswerLet $y=\cos ^{-1}\left(\sqrt{\frac{1+x}{2}}\right)$
Differentiate w.r.t.x.
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left[\cos ^{-1}\left(\sqrt{\frac{1+x}{2}}\right)\right] \\
& =-\frac{1}{\sqrt{1-\left(\sqrt{\frac{1+x}{2}}\right)^2}} \cdot \frac{d}{d x}\left(\sqrt{\frac{1+x}{2}}\right) \\
& =-\frac{1}{\sqrt{1-\frac{1+x}{2}}} \times \frac{1}{2 \sqrt{\frac{1+x}{2}}} \times \frac{d}{d x}\left(\frac{1+x}{2}\right) \\
& =-\frac{\sqrt{2}}{\sqrt{1-x}} \times \frac{1}{\sqrt{2} \sqrt{1+x}} \times \frac{1}{2} \\
\therefore \frac{d y}{d x} & =-\frac{1}{2 \sqrt{1-x^2}}
\end{aligned}
$
View full question & answer→Question 242 Marks
Differentiate the following w. r. t. x.$\cos ^{-1}\left(2 x^2-x\right)$
AnswerLet $y=\cos ^{-1}\left(2 x^2-x\right)$
Hence $\cos y=2 x^2-x$
Differentiate w.r.t. $x$.
$
-\sin y \cdot \frac{d y}{d x}=4 x-1
$
$
\begin{aligned}
\frac{d y}{d x} & =\frac{1-4 x}{\sin y}=\frac{1-4 x}{\sqrt{1-\cos ^2 y}} \\
\therefore \frac{d y}{d x} & =\frac{1-4 x}{\sqrt{1-x^2(2 x-1)^2}} \quad \ldots \text { from (I) }
\end{aligned}
$
... from (I)
Alternate Method :
$
\begin{aligned}
& \text { If } y=\cos ^{-1}\left(2 x^2-x\right) \\
& \text { Differentiate w.r.t.x. } \\
& \begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left(\cos ^{-1}\left(2 x^2-x\right)\right) \\
& =\frac{-1}{\sqrt{1-\left(2 x^2-x\right)^2}} \cdot \frac{d}{d x}\left(2 x^2-x\right) \\
& =\frac{-1}{\sqrt{1-x^2(2 x-1)^2}} \cdot(4 x-1) \\
\therefore \quad \frac{d y}{d x} & =\frac{1-4 x}{\sqrt{1-x^2(2 x-1)^2}}
\end{aligned}
\end{aligned}
$
View full question & answer→Question 252 Marks
Using derivative prove that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$.
AnswerLet $f(x)=\sin ^{-1} x+\cos ^{-1} x \quad \ldots .(I)$
We have to prove that $f(x)=\frac{\pi}{2}$
Differentiate (I) w.r.t.x
$
\begin{aligned}
\frac{d}{d x}[f(x)] & =\frac{d}{d x}\left[\sin ^{-1} x+\cos ^{-1} x\right] \\
f^{\prime}(x) & =\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}=0
\end{aligned}
$
$f^{\prime}(x)=0 \Rightarrow f(x)$ is a constant function.
Let $f(x)=c$. For any value of $x, f(x)$ must be $c$ only. So conveniently we can choose $x=0$,
$\therefore \quad$ from (I) we get,
$
f(0)=\sin ^{-1}(0)+\cos ^{-1}(0)=0+\frac{\pi}{2}=\frac{\pi}{2} \Rightarrow c=\frac{\pi}{2} \therefore f(x)=\frac{\pi}{2}
$
Hence, $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$.
View full question & answer→Question 262 Marks
Find the derivative of the function $y=f(x)$ using the derivative of the inverse function $x=f^{-1}(y)$ in the following$y=\sqrt{1+\sqrt{x}}$
Answer$
y=\sqrt{1+\sqrt{x}}
$
We first find the inverse of the function $y=f(x)$, i.e. $x$ in term of $y$.
$
\begin{aligned}
& y^2=1+\sqrt{x} \text { i.e. } \sqrt{x}=y^2-1, \therefore x=f^{-1}(y)=\left(y^2-1\right)^2 \\
& \begin{aligned}
\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}=\frac{1}{\frac{d}{d y}\left[\left(y^2-1\right)^2\right]} & =\frac{1}{2\left(y^2-1\right) \frac{d}{d y}\left(y^2-1\right)} \\
= & \frac{1}{2\left(y^2-1\right)(2 y)}=\frac{1}{4 \sqrt{1+\sqrt{x}}\left[(\sqrt{1+\sqrt{x}})^2-1\right]} \\
= & \frac{1}{4 \sqrt{1+\sqrt{x}}(1+\sqrt{x}-1)}=\frac{1}{4 \sqrt{x} \sqrt{1+\sqrt{x}}}
\end{aligned}
\end{aligned}
$
View full question & answer→Question 272 Marks
If $F(x)=G\{3 G[5 G(x)]\}, G(0)=0$ and $G^{\prime}(0)=3$, find $F^{\prime}(0)$.
AnswerGiven that: $F(x)=G\{3 G[5 G(x)]\}$
Differentiate w.r.t.x
$
\begin{aligned}
F^{\prime}(x) & =\frac{d}{d x} G\{3 G[5 G(x)]\} \\
& =G^{\prime}\{3 G[5 G(x)]\} 3 \cdot \frac{d}{d x}[G[5 G(x)]] \\
& =G^{\prime}\{3 G[5 G(x)]\} 3 \cdot G^{\prime}[5 G(x)] 5 \cdot \frac{d}{d x}[G(x)]
\end{aligned}
$
$
F^{\prime}(x)=15 \cdot G^{\prime}\{3 G[5 G(x)]\} G^{\prime}[5 G(x)] G^{\prime}(x)
$
For $x=0$, we get
$
\begin{array}{rlr}
F^{\prime}(0) & =15 \cdot G^{\prime}\{3 G[5 G(0)]\} G^{\prime}[5 G(0)] G^{\prime}(0) \\
& =15 \cdot G^{\prime}[3 G(0)] G^{\prime}(0) \cdot(3) & {\left[\because G(0)=0 \text { and } G^{\prime}(0)=3\right]} \\
& =15 \cdot G^{\prime}(0)(3)(3)=15 \cdot(3)(3)(3)=405
\end{array}
$
View full question & answer→Question 282 Marks
If $f(x)=\sqrt{7 g(x)-3}, g(3)=4$ and $g^{\prime}(3)=5$, find $f^{\prime}(3)$.
Answer$\begin{aligned} & \text { Given that: } f(x)=\sqrt{7 g(x)-3} \\ & \text { Differentiate } \text {.r.r.t. } \\ & f^{\prime}(x)=\frac{d}{d x}(\sqrt{7 g(x)-3})=\frac{1}{2 \sqrt{7 g(x)-3}} \frac{d}{d x}[7 g(x)-3] \\ & \therefore \quad f^{\prime}(x)=\frac{7 g^{\prime}(x)}{2 \sqrt{7 g(x)-3}} \\ & \text { For } x=3 \text {, we get } \\ & f^{\prime}(3)=\frac{7 g^{\prime}(3)}{2 \sqrt{7 g(3)-3}}=\frac{35}{2(5)}=\frac{7}{2} \quad\left[\text { Since } g(3)=4 \text { and } g^{\prime}(3)=5\right]\end{aligned}$
View full question & answer→Question 292 Marks
Differentiate $\tan ^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ w.r.t $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$
View full question & answer→Question 302 Marks
Differentiate $\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ w.r.t $\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
View full question & answer→Question 312 Marks
If $x=\sin ^{-1}\left(e^t\right), y=\sqrt{1-e^{2 t}}$, show that $\sin x+\frac{d y}{d x}=0$
View full question & answer→Question 322 Marks
Find $\frac{d y}{d x}$ if
x = a(1 – cos θ), y = b(θ – sin θ)
View full question & answer→Question 332 Marks
Find $\frac{d y}{d x}$ if
x = sin θ, y = tan θ
View full question & answer→Question 342 Marks
Find $\frac{d y}{d x}$ if
x = a cot θ, y = b cosec θ
View full question & answer→Question 352 Marks
Find $\frac{d y}{d x}$ if$x=a t^2, y=2 a t$
View full question & answer→Question 362 Marks
If $y=\sqrt{\log x+\sqrt{\log x+\sqrt{\log x+\ldots \infty}}}$, then show that $\frac{d y}{d x}=\frac{1}{x(2 y-1)}$
View full question & answer→Question 372 Marks
If $y=\sqrt{\cos x+\sqrt{\cos x+\sqrt{\cos x+\ldots \infty}}}$, then show that $\frac{d y}{d x}=\frac{\sin x}{1-2 y}$
View full question & answer→Question 382 Marks
If $\sin ^{-1}\left(\frac{x^5-y^5}{x^5+y^5}\right)=\frac{\pi}{6}$, show that $\frac{d y}{d x}=\frac{x^4}{3 y^4}$
View full question & answer→Question 392 Marks
If $e ^{ x }+ e ^y= e ^{ x + y }$, then show that $\frac{d y}{d x}=-e^{y-x}$
View full question & answer→Question 402 Marks
If $\log _{10}\left(\frac{x^2-y^3}{x^3+y^3}\right)=2$, show that $\frac{d y}{d x}=-\frac{99 x^2}{101 y^2}$
View full question & answer→Question 412 Marks
Find $\frac{d y}{d x}$ if$x \sqrt{x}+y \sqrt{y}=a \sqrt{a}$
View full question & answer→Question 422 Marks
Find $\frac{d y}{d x}$ if$\sqrt{x}+\sqrt{y}=\sqrt{a}$
View full question & answer→Question 432 Marks
Diffrentiate the following w. r. t. x. $\tan ^{-1}\left(\frac{8 x}{1-15 x^2}\right)$
View full question & answer→Question 442 Marks
Diffrentiate the following w. r. t. x$\cot ^{-1}\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)$
View full question & answer→Question 452 Marks
Diffrentiate the following w. r. t. x$\tan ^{-1}\left(\frac{2 x^{\frac{5}{2}}}{1-x^5}\right)$
View full question & answer→Question 462 Marks
Diffrentiate the following w. r. t. x$\sin ^{-1}\left(\frac{1-x^3}{1+x^3}\right)$
View full question & answer→Question 472 Marks
Diffrentiate the following w. r. t. x$\sin ^{-1}\left(\frac{1-25 x^2}{1+25 x^2}\right)$
View full question & answer→Question 482 Marks
Diffrentiate the following w. r. t. x$\sin ^{-1}\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)$
View full question & answer→Question 492 Marks
Diffrentiate the following w. r. t. x$\operatorname{CoS}^{-1}\left(\frac{1-9^x}{1+9^x}\right)$
View full question & answer→Question 502 Marks
Diffrentiate the following w. r. t. x$\cos ^{-1}\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)$
View full question & answer→Question 512 Marks
Diffrentiate the following w. r. t. x$\cos ^{-1}\left(3 x-4 x^3\right)$
View full question & answer→Question 522 Marks
Diffrentiate the following w. r. t. x$\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$
View full question & answer→Question 532 Marks
Diffrentiate the following w. r. t. x$\sin ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
View full question & answer→Question 542 Marks
Diffrentiate the following w. r. t. x$\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)$
View full question & answer→Question 552 Marks
Diffrentiate the following w. r. t. x. $\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)$
View full question & answer→Question 562 Marks
Diffrentiate the following w. r. t. x.$\operatorname{cosec}^{-1}\left(\frac{1}{\cos \left(5^x\right)}\right)$
AnswerLet $\begin{aligned} y & =\operatorname{cosec}^{-1}\left[\frac{1}{\cos \left(5^x\right)}\right] \\ & =\operatorname{cosec}^{-1}\left[\sec \left(5^x\right)\right] \\ & =\operatorname{cosec}^{-1}\left[\operatorname{cosec}\left(\frac{\pi}{2}-5^x\right)\right] \\ & =\frac{\pi}{2}-5^x\end{aligned}$Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(\frac{\pi}{2}-5^x\right) \\ & =\frac{d}{d x}\left(\frac{\pi}{2}\right)-\frac{d}{d x}\left(5^x\right) \\ & =0-5^x \cdot \log 5 \\ & =-5^x \cdot \log 5\end{aligned}$
View full question & answer→Question 572 Marks
Find the derivative of the inverse of the following functions, and also fid their value at the points indicated against them :
If $f(x)=x^3+x-2$, find $\left(f^{-1}\right)^{\prime}(0)$
Question is modified.
If $f(x)=x^3+x-2$, find $\left(f^{-1}\right)^{\prime}(-2)$.
Answer$f(x)=x^3+x-2$
Differentiating w.r.t. $x$, we get
$ f^{\prime}(x) & =\frac{d}{d x}\left(x^3+x-2\right)$
$ =3 x^2+1-0=3 x^2+1 $
We know that
$\left(f^{-1}\right)^{\prime}(y)=\frac{1}{f^{\prime}(x)}$
From (1), $y=f(x)=-2$, when $x=0$
$ \therefore \text { from (2), }\left(f^{-1}\right)^{\prime}(-2)=\frac{1}{f^{\prime}(0)}=\frac{1}{\left(3 x^2+1\right)_{\text {at } x=0}}$
$=\frac{1}{3(0)+1}=1 . $
View full question & answer→Question 582 Marks
Find the derivative of the inverse of the following functions, and also fid their value at the points indicated against them :
$y=e^x+3 x+2$, at $x=0$
Answer$y=e^x+3 x+2$Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left( e ^{ x }+3 x +2\right)$
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}{e^x+3}$
$\text { At } x=0, \frac{d x}{d y}=\frac{1}{\left(e^x+3\right)_{\text {at } x=0}}$
$=\frac{1}{e^0+3}=\frac{1}{1+3}=\frac{1}{4} \text {. }$
View full question & answer→Question 592 Marks
Find the derivative of the inverse of the following functions, and also fid their value at the points indicated against them.
$y=x^5+2 x^3+3 x_r$ at $x=1$
Answer$y=x^5+2 x^3+3 x$Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(x^5+2 x^3+3 x\right)$
$=5 x^4+2 \times 3 x^2+3 \times 1$
$=5 x^4+6 x^2+3$
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}{5 x^4+6 x^2+3}$
At $x=1, \frac{d x}{d y}=\frac{1}{\left(5 x^4+6 x^2+3\right)_{\text {at } x=1}}$
$=\frac{1}{5(1)^4+6(1)^2+3}$
$=\frac{1}{5+6+3}=\frac{1}{14} .$
View full question & answer→Question 602 Marks
Find the derivative of the inverse function of the following
y = x logx
Answery = x logx Differentiating w.r.t. x, we get $\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(x \log x) \\ & =x \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x) \\ & =x \times \frac{1}{x}+(\log x) \times 1 \\ & =1+\log x\end{aligned}$
The derivative of inverse function of $y=f(x)$ is given by
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(x \log x) \\ & =x \frac{d}{d x}(\log x)+(\log x) \cdot \frac{d}{d x}(x) \\ & =x \times \frac{1}{x}+(\log x) \times 1 \\ & =1+\log x\end{aligned}$
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}{1+\log x}$
View full question & answer→Question 612 Marks
Find the derivative of the inverse function of the following$y=x^2+\log x$
Answer$y=x^2+\log x$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(x^2+\log x\right) \\ & =\frac{d}{d x}\left(x^2\right)+\frac{d}{d x}(\log x) \\ & =2 x+\frac{1}{x}=\frac{2 x^2+1}{x}\end{aligned}$
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{2 x^2+1}{x}\right)}=\frac{x}{2 x^2+1}$
View full question & answer→Question 622 Marks
Find the derivative of the inverse function of the following
y = x·7x
Answery = x·7x Differentiating w.r.t. x, we get$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(x \cdot 7^x\right) \\ & =x \frac{d}{d x}\left(7^x\right)+7^x \frac{d}{d x}(x) \\ & =x \cdot 7^x \log 7+7^x \times 1 \\ & =7^x(x \log 7+1)\end{aligned}$
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}{7^x(x \log 7+1)}$
View full question & answer→Question 632 Marks
Find the derivative of the inverse function of the following
y = x cos x
Answery = x cos x Differentiating w.r.t. x, we get$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(x \cos x) \\ & =x \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}(x) \\ & =x(-\sin x)+\cos x \times 1 \\ & =\cos x-x \sin x\end{aligned}$
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-x \sin x}$
View full question & answer→Question 642 Marks
Find the derivative of the inverse function of the following$y=x^2 \cdot e^x$
Answer$y=x^2 \cdot e^x$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(x^2 \cdot e^x\right) \\ & =x^2 \frac{d}{d x}\left(e^x\right)+e^x \frac{d}{d x}\left(x^2\right) \\ & =x^2 \cdot e^x+e^x \times 2 x \\ & =x e^x(x+2)\end{aligned}$
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}{x e^x(x+2)}$
View full question & answer→Question 652 Marks
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f-1( y) in the following $y=\log _2\left(\frac{x}{2}\right)$
Answer$y=\log _2\left(\frac{x}{2}\right) \ldots(1)$We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & \frac{x}{2}=2^y \therefore x=2 \cdot 2^y=2^{y+1} \\ & \therefore x=f^{-1}(y)=2^{y+1}\end{aligned}$
$\begin{aligned} \therefore \frac{d x}{d y} & =\frac{d}{d y}\left(2^{y+1}\right) \\ & =2^{y+1} \cdot \log 2 \cdot \frac{d}{d y}(y+1) \\ & =2^{y+1} \cdot \log 2 \cdot(1+0) \\ & =2^{y+1} \cdot \log 2=2^{\log _2\left(\frac{x}{2}\right)+1} \cdot \log 2 \ldots \text { [Вy (1)] }\end{aligned}$
$\begin{aligned} & =2^{\log _2\left(\frac{x}{2}\right)+\log _2 2} \cdot \log 2 \\ & =2^{\log _2\left(\frac{x}{2} \times 2\right)} \cdot \log 2=2^{\log _2 x} \cdot \log 2 \\ & =x \log 2 \quad \ldots\left[\because a^{\log _5 x}=x\right]\end{aligned}$
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{x \log 2}$
View full question & answer→Question 662 Marks
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f-1( y) in the following $y=e^{2 x-3}$
Answer$y=e^{2 x-3} \ldots(1)$We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1), 2x – 3 = log y ∴ 2x = log y + 3
$\begin{aligned} \therefore x & =f^{-1}(y)=\frac{1}{2}(\log y+3) \\ \therefore \frac{d x}{d y} & =\frac{1}{2} \frac{d}{d y}(\log y+3) \\ & =\frac{1}{2}\left(\frac{1}{y}+0\right)=\frac{1}{2 y}\end{aligned}$
$=\frac{1}{2 e^{2 x-3}}$
... [By (1)]
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{\left(\frac{1}{2 e^{2 x-3}}\right)}=2 e^{2 x-3}$
View full question & answer→Question 672 Marks
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f-1( y) in the following $y=e^x-3$
Answer$y=e^x-3 \ldots(1)$We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1), ex = y + 3 ∴ x = log(y + 3) ∴ x = f-1(y) = log(y + 3)
$\begin{aligned} \therefore \frac{d x}{d y} & =\frac{d}{d y}[\log (y+3)] \\ & =\frac{1}{y+3} \cdot \frac{d}{d y}(y+3) \\ & =\frac{1}{y+3} \cdot(1+0)=\frac{1}{y+3} \\ & =\frac{1}{e^x-3+3} \\ & =\frac{1}{e^x}\end{aligned}$
... [By (1)]
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{\left(\frac{1}{e^x}\right)}=e^x$.
View full question & answer→Question 682 Marks
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f-1( y) in the following y = 2x + 3
Answery = 2x + 3 ….(1) We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),$\begin{aligned} & 2 x=y-3 \quad \therefore x=\frac{y-3}{2} \\ & \therefore x=f^{-1}(y)=\frac{y-3}{2}\end{aligned}$
$\therefore \frac{d x}{d y}=\frac{1}{2} \frac{d}{d y}(y-3)$
$\begin{aligned} \quad & \frac{1}{2}(1-0)=\frac{1}{2} \\ \therefore \frac{d y}{d x} & =\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{\left(\frac{1}{2}\right)}=2\end{aligned}$
View full question & answer→Question 692 Marks
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f-1( y) in the following y = log (2x – 1)
Answery = log (2x – 1) …(1)
We have to find the inverse function of y = f(x), i.e. x in terms of y.
From (1),y = 2x + 3 ….(1) We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & 2 x-1=e^y \quad \therefore 2 x=e^y+1 \\ & \therefore x=f^{-1}(y)=\frac{1}{2}\left(e^y+1\right) \\ & \therefore \frac{d x}{d y}=\frac{1}{2} \frac{d}{d y}\left(e^y+1\right) \\ & \quad=\frac{1}{2}\left(e^y+0\right)=\frac{1}{2} e^y\end{aligned}$
$=\frac{1}{2} e^{\log (2 x-1)} \quad \ldots$ [By (1)]
$=\frac{1}{2}(2 x-1) \quad \ldots\left[\because e^{\log x}=x\right]$
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{2}{2 x-1}$
View full question & answer→Question 702 Marks
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f-1( y) in the following $y=\sqrt[3]{x-2}$
Answer$y=\sqrt[3]{x-2} \ldots(1)$We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & y^3=x-2 \quad \therefore x=y^3+2 \\ & \therefore x=f^{-1}(y)=y^3+2 \\ & \therefore \frac{d x}{d y}=\frac{d}{d y}\left(y^3+2\right) \\ & \quad=3 y^2+0=3 y^2\end{aligned}$
$=3(\sqrt[3]{(x-2)})^2 \quad \ldots[$ By (1)]
$=3(x-2)^3=3 \cdot\left(\sqrt[3]{(x-2)^2}\right)$
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d Y}\right)}=\frac{1}{\left.3 \sqrt[3]{(x-2)^2}\right)}, x>2$
View full question & answer→Question 712 Marks
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f-1( y) in the following $y=\sqrt{2-\sqrt{x}}$
Answer$y =\sqrt{2-\sqrt{x}} \ldots$ (1)We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & y^2=2-\sqrt{x} \quad \therefore \sqrt{x}=2-y^2 \\ & \therefore x=\left(2-y^2\right)^2 \\ & \therefore x=f^{-1}(y)=\left(2-y^2\right)^2 \\ & \therefore \frac{d x}{d y}=\frac{d}{d y}\left(2-y^2\right)^2 \\ & \quad=2\left(2-y^2\right) \cdot \frac{d}{d y}\left(2-y^2\right) \\ & \quad=2\left(2-y^2\right) \cdot(0-2 y) \\ & \quad=-4 y\left(2-y^2\right)\end{aligned}$
$\begin{aligned} & =-4 \sqrt{2-\sqrt{x}}(2-2+\sqrt{x}) \quad \ldots \text { [By (1)] } \\ & =-4 \sqrt{x} \sqrt{2-\sqrt{x}}\end{aligned}$
$\therefore \frac{d y}{d x}=\frac{1}{\left(\frac{d x}{d y}\right)}=-\frac{1}{4 \sqrt{x} \sqrt{2-\sqrt{x}}}$
View full question & answer→Question 722 Marks
Find the derivative of the function y = f(x) using the derivative of the inverse function x = f-1( y) in the following $y=\sqrt{x}$
Answer$y=\sqrt{x} \ldots(1)$We have to find the inverse function of y = f(x), i.e. x in terms of y. From (1),
$\begin{aligned} & y ^2= x \therefore x = y ^2 \\ & \therefore x=f^{-1}(y)=y^2 \\ & \therefore \frac{d x}{d y}=\frac{d}{d y}\left(y^2\right)=2 y\end{aligned}$
$\begin{aligned} & =2 \sqrt{x} \\ \therefore \frac{d y}{d x} & =\frac{1}{\left(\frac{d x}{d y}\right)}=\frac{1}{2 \sqrt{x}} .\end{aligned}$
$\ldots$ [By (1)]
View full question & answer→Question 732 Marks
"Let $f(x)=x^2+5$ and $g(x)=e^x+3$ then
$f[g(x)]=$ and $g[f(x)]=$
Now $f^{\prime}(x)=$ and $g^{\prime}(x)=$
The derivative off $[g(x)]$ w.r. t. $x$ in terms of $f$ and $g$ is
Therefore $\frac{d}{d x}[f[g(x)]]=$ and $\left[\frac{d}{d x}[f[g(x)]]\right]_x=0=$
The derivative of $g[f(x)]$ w.r. t. $x$ in terms of $f$ and $g$ is
Therefore $\frac{d}{d x}[ g [ f ( x )]]=\ldots . . . .$. and $\left[\frac{d}{d x}[ g [ f ( x )]]\right]_{ x =1}=$
Hint basket: $\left\{f^{\prime}[g(x)] \cdot g^{\prime}(x), 2 e^{2 x}+6 e^x, 8, g^{\prime}[f(x)] \cdot f^{\prime}(x), 2 x e^{x 2+5},-2 e^6, e^{2 x}+6 e^x+14, e^{x 2}+\right.$ $\left.5+3,2 x, e^x\right\}$
Answer$\begin{aligned} & f[g(x)]=e^{2 x}+6 e^x+14 \\ & g[f(x)]=e^{x 2+5}+3 \\ & f^{\prime}(x)=2 x, g^{\prime} f(x)=e^x\end{aligned}$The derivative of f[g(x)] w.r.t. x in terms of and g is f'[g(x)]∙g'(x)
$\therefore \frac{d}{d x}\{ f [ g ( x )]\}=2 e ^{2 x }+6 e ^{ x }$ and $\frac{d}{d x}\{ f [ g ( x )]\}_{ x }=0=8$
The derivative of g[f(x)] w.r.t. x in terms of f and g is g’f(x)]∙f'(x).
$\begin{aligned} & \therefore \frac{d}{d x}\left\{ g [( f ( x )]\}=2 xe ^{ x 2+5} \text { and }\right. \\ & \frac{d}{d x}\left\{ g [( f ( x )]\}_{ x }=-1\right. \\ & =-2 e ^6\end{aligned}$
View full question & answer→Question 742 Marks
Differentiate the following w.r.t. x :$\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^5$
AnswerLet $y =\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^5$Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^5$
$\begin{aligned} & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot \frac{d}{d x}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right) \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot\left[\frac{d}{d x}(3 x-5)^{\frac{1}{2}}-\frac{d}{d x}(3 x-5)^{-\frac{1}{2}}\right] \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \times \\ & {\left[\frac{1}{2}(3 x-5)^{-\frac{1}{2}} \cdot \frac{d}{d x}(3 x-5)-\left(-\frac{1}{2}\right)(3 x-5)^{-\frac{3}{2}} \cdot \frac{d}{d x}(3 x-5)\right]} \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \times \\ & {\left[\frac{1}{2 \sqrt{3 x-5}} \cdot(3 \times 1-0)+\frac{1}{2(3 x-5)^{\frac{3}{2}}} \cdot(3 \times 1-0)\right]} \\ & =5\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot\left[\frac{3}{2 \sqrt{3 x-5}}+\frac{3}{2(3 x-5)^{\frac{3}{2}}}\right] \\ & =\frac{15}{2}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \cdot\left[\frac{3 x-5+1}{(3 x-5)^{\frac{3}{2}}}\right] \\ & =\frac{15(3 x-4)}{2(3 x-5)^{\frac{3}{2}}}\left(\sqrt{3 x-5}-\frac{1}{\sqrt{3 x-5}}\right)^4 \text {. } \\ & \end{aligned}$
View full question & answer→Question 752 Marks
Differentiate the following w.r.t. x :$\frac{3}{5 \sqrt[3]{\left(2 x^2-7 x-5\right)^5}}$
AnswerLet $y =\frac{3}{5 \sqrt[3]{\left(2 x^2-7 x-5\right)^5}}$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{3}{5} \frac{d}{d x}\left(2 x^2-7 x-5\right)^{-\frac{5}{3}} \\ & =\frac{3}{5} \times\left(-\frac{5}{3}\right)\left(2 x^2-7 x-5\right)^{-\frac{5}{3}-1} \cdot \frac{d}{d x}\left(2 x^2-7 x-5\right) \\ & =-\left(2 x^2-7 x-5\right)^{-\frac{8}{3}} \cdot(2 \times 2 x-7 \times 1-0) \\ & =-\frac{4 x-7}{\left(2 x^2-7 x-5\right)^{\frac{8}{3}}} \cdot\end{aligned}$
View full question & answer→Question 762 Marks
Differentiate the following w.r.t. x :$\sqrt{x^2+\sqrt{x^2+1}}$
AnswerLet $y=\sqrt{x^2+\sqrt{x^2+1}}$Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(x^2+\sqrt{x^2+1}\right)^{\frac{1}{2}}$
$\begin{aligned} & =\frac{1}{2}\left(x^2+\sqrt{x^2+1}\right)^{-\frac{1}{2}} \cdot \frac{d}{d x}\left(x^2+\sqrt{x^2+1}\right) \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[\frac{d}{d x}\left(x^2\right)+\frac{d}{d x}\left(\sqrt{x^2+1}\right)\right] \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[2 x+\frac{1}{2 \sqrt{x^2+1}} \cdot \frac{d}{d x}\left(x^2+1\right)\right] \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[2 x+\frac{1}{2 \sqrt{x^2+1}}(2 x+0)\right] \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[2 x+\frac{x}{\sqrt{x^2+1}}\right] \\ & =\frac{1}{2 \sqrt{x^2+\sqrt{x^2+1}}} \cdot\left[\frac{2 x \sqrt{x^2+1}+x}{\sqrt{x^2+1}}\right] \\ & 2 \sqrt{x\left(2 \sqrt{x^2+1} \cdot \sqrt{x^2+1}+1\right)} \cdot \sqrt{x^2+1} \cdot\end{aligned}$
View full question & answer→Question 772 Marks
Differentiate the following w.r.t. x :$\sqrt{x^2+4 x-7}$
Answer$y=\sqrt{x^2+4 x-7}\left[\sqrt{x}=\frac{1}{2 \sqrt{x}}\right]$Differentiating w.r.t. $x$, we get
$\begin{aligned} & \frac{ dy }{ dx }=\frac{1}{2 \sqrt{x^2+4 x-7}} \cdot \frac{ d }{ dx }\left(x^2+4 x-7\right) \\ & =\frac{1}{2 \sqrt{x^2+4 x-7}}\left(\frac{ d }{ dx } x^2+\frac{ d }{ dx } 4 x-\frac{ d }{ dx } 7\right) \\ & =\frac{1}{2 \sqrt{x^2+4 x-7}} \cdot(2 x+4-0) \\ & =\frac{2(x+2)}{2 \sqrt{x^2+4 x-7}} \\ & =\frac{(x+2)}{\sqrt{x^2+4 x-7}} .\end{aligned}$
View full question & answer→Question 782 Marks
Differentiate the following w.r.t. x :$\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}$
AnswerLet $y=\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}}$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}} \\ & =\frac{5}{2}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{5}{2}-1} \times \frac{d}{d x}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right) \\ & =\frac{5}{2}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{3}{2}} \times\left(2 \times \frac{3}{2} x^{\frac{3}{2}-1}-3 \times \frac{4}{3} x^{\frac{4}{3}-1}-0\right) \\ & =\frac{5}{2}\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{3}{2}}\left(3 x^{\frac{1}{2}}-4 x^{\frac{1}{3}}\right) \\ & =\frac{5}{2}(3 \sqrt{x}-4 \sqrt[3]{x})\left(2 x^{\frac{3}{2}}-3 x^{\frac{4}{3}}-5\right)^{\frac{3}{2}}\end{aligned}$
View full question & answer→Question 792 Marks
Differentiate the following w.r.t. x :$\left(x^3-2 x-1\right)^5$
AnswerMethod 1: Let $y=\left(x^3-2 x-1\right)^5$ Put $u=x^3-2 x-1$. Then $y=u^5$$\therefore \frac{d y}{d u}=\frac{d}{d u}\left(u^5\right)=5 u^4$
$\begin{aligned} & =5\left(x^3-2 x-1\right)^4 \\ \text { and } \frac{d u}{d x} & =\frac{d}{d x}\left(x^3-2 x-1\right) \\ & =3 x^2-2 \times 1-0=3 x^2-2 \\ \therefore \frac{d y}{d x} & =\frac{d y}{d u} \times \frac{d u}{d x} \\ & =5\left(x^3-2 x-1\right)^4\left(3 x^2-2\right) \\ & =5\left(3 x^2-2\right)\left(x^3-2 x-1\right)^4 .\end{aligned}$
Method 2: Let $y=\left(x^3-2 x-1\right)^5$ Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left(x^3-2 x-1\right)^5 \\ & =5\left(x^3-2 x-1\right)^4 \times \frac{d}{d x}\left(x^3-2 x-1\right) \\ & =5\left(x^3-2 x-1\right)^4 \times\left(3 x^2-2 \times 1-0\right) \\ & =5\left(3 x^2-2\right)\left(x^3-2 x-1\right)^4\end{aligned}$
View full question & answer→Question 802 Marks
Assume that $f ^{\prime}(3)=-1, g ^{\prime}(2)=5, g(2)=3$ and $y = f [ g ( x )]$ then $\left[\frac{d y}{d x}\right]_{x=2}=$ ?
Answer$\begin{aligned} & y = f [ g ( x )] \\ & \begin{aligned} \therefore \frac{d y}{d x} & =\frac{d}{d x}\{[ g ( x )]\} \\ & =f^{\prime}[g(x)] \cdot \frac{d}{d x}[g(x)] \\ & =f^{\prime}[g(x)] \cdot g^{\prime}(x)\end{aligned}\end{aligned}$$\begin{array}{rlr}\therefore\left[\frac{d y}{d x}\right]_{x=2} & =f^{\prime}[g(2)] \cdot g^{\prime}(2) & \\ & =f^{\prime}(3) \cdot g^{\prime}(2) & \ldots[\because g(2)=3] \\ & =-1 \times 5 & \ldots \text { (Given) } \\ & =-5 .\end{array}$
View full question & answer→Question 812 Marks
Diffrentiate the following w.r.t.x$\sin ^2 x^2-\cos ^2 x^2$
AnswerLet $y=\sin ^2 x^2-\cos ^2 x^2$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\sin ^2 x^2-\cos ^2 x^2\right] \\ & =\frac{d}{d x}\left(\sin x^2\right)^2-\frac{d}{d x}\left(\cos x^2\right)^2 \\ & =2 \sin x^2 \cdot \frac{d}{d x}\left(\sin x^2\right)-2 \cos x^2 \cdot \frac{d}{d x}\left(\cos x^2\right) \\ & =2 \sin x^2 \cdot \cos x^2 \cdot \frac{d}{d x}\left(x^2\right)-2 \cos x^2 \cdot\left(-\sin x^2\right) \cdot \frac{d}{d x}\left(x^2\right) \\ & =2 \sin x^2 \cdot \cos x^2 \times 2 x+2 \sin x^2 \cdot \cos x^2 \times 2 x \\ & =4 x\left(2 \sin x^2 \cdot \cos x^2\right) \\ & =4 x \sin \left(2 x^2\right) .\end{aligned}$
View full question & answer→Question 822 Marks
Diffrentiate the following w.r.t.x$[\log \{\log (\log x)\}]^2$
Answerlet $y=[\log \{\log (\log x)\}]^2$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\log \{\log (\log x)\}]^2 \\ & =2 \cdot \log \{\log (\log x)\} \times \frac{d}{d x}[\log \{\log (\log x)\}]\end{aligned}$
$\begin{aligned} & =2 \cdot \log \{\log (\log x)\} \times \frac{1}{\log (\log x)} \cdot \frac{d}{d x}[\log (\log x)] \\ & =2 \cdot \log \{\log (\log x)\} \times \frac{1}{\log (\log x)} \times \frac{1}{\log x} \times \frac{d}{d x}(\log x) \\ & =2 \cdot \log \{\log (\log x)\} \times \frac{1}{\log (\log x)} \times \frac{1}{\log x} \times \frac{1}{x} \\ & =2 \cdot\left[\frac{\log \{\log (\log x)\}}{x \cdot \log x \cdot \log (\log x)}\right] .\end{aligned}$
View full question & answer→Question 832 Marks
Diffrentiate the following w.r.t.x$\log _{e^2}(\log x)$
AnswerLet $y =\log _{e^2(\log x)}=\frac{\log (\log x)}{\log e^2}$$=\frac{\log (\log x)}{2 \log e}=\frac{\log (\log x)}{2} \quad \ldots[\because \log e=1]$
Differentiating w.r.t. $x$, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{1}{2} \frac{d}{d x}[\log (\log x)] \\ & =\frac{1}{2} \times \frac{1}{\log x} \cdot \frac{d}{d x}(\log x) \\ = & \frac{1}{2 \log x} \times \frac{1}{x}=\frac{1}{2 x \log x}\end{aligned}$
View full question & answer→Question 842 Marks
Diffrentiate the following w.r.t.x$\log \left[\sec \left( e ^{\times 2}\right)\right]$
AnswerLet $y=\log \left[\sec \left(e^{x 2}\right)\right]$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\log \left(\sec e^{x^2}\right)\right] \\ & =\frac{1}{\sec \left(e^{x^2}\right)} \cdot \frac{d}{d x}\left[\sec \left(e^{x^2}\right)\right] \\ & =\frac{1}{\sec \left(e^{x^2}\right)} \cdot \sec \left(e^{x^2}\right) \tan \left(e^{x^2}\right) \cdot \frac{d}{d x}\left(e^{x^2}\right) \\ & =\tan \left(e^{x^2}\right) \cdot e^{x^2} \cdot \frac{d}{d x}\left(x^2\right) \\ & =\tan \left(e^{x^2}\right) \cdot e^{x^2} \cdot 2 x \\ & =2 x \cdot e^{x^2} \tan \left(e^{x^2}\right) .\end{aligned}$
View full question & answer→Question 852 Marks
Diffrentiate the following w.r.t.x$\sin \sqrt{\sin \sqrt{x}}$
AnswerLet $y =\sin \sqrt{\sin \sqrt{x}}$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(\sin \sqrt{\sin \sqrt{x}}) \\ & =\cos \sqrt{\sin \sqrt{x}} \cdot \frac{d}{d x}(\sqrt{\sin \sqrt{x}}) \\ & =\cos \sqrt{\sin \sqrt{x}} \times \frac{1}{2 \sqrt{\sin \sqrt{x}}} \cdot \frac{d}{d x}(\sin \sqrt{x}) \\ & =\frac{\cos \sqrt{\sin \sqrt{x}}}{2 \sqrt{\sin \sqrt{x}}} \times \cos \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x})\end{aligned}$
$\begin{aligned} & =\frac{\cos \sqrt{\sin \sqrt{x}} \cdot \cos \sqrt{x}}{2 \sqrt{\sin \sqrt{x}}} \times \frac{1}{2 \sqrt{x}} \\ & =\frac{\cos \sqrt{\sin \sqrt{x}} \cdot \cos \sqrt{x}}{4 \sqrt{x} \cdot \sqrt{\sin \sqrt{x}}} .\end{aligned}$
View full question & answer→Question 862 Marks
Diffrentiate the following w.r.t.x$e^{\log [(\log x) 2-\log x 2]}$
Answer$\begin{aligned} & \text { Let } y=e^{\log [(\log x) 2-\log x 2]} \\ & =(\log x)^2-\log x^2 \ldots\left[\because e^{\log x}=x\right]\end{aligned}$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[(\log x)^2-2 \log x\right] \\ & =\frac{d}{d x}(\log x)^2-2 \frac{d}{d x}(\log x) \\ & =2 \log x \cdot \frac{d}{d x}(\log x)-2 \times \frac{1}{x} \\ & =2 \log x \times \frac{1}{x}-\frac{2}{x} \\ & =\frac{2 \log x}{x}-\frac{2}{x} .\end{aligned}$
View full question & answer→Question 872 Marks
Diffrentiate the following w.r.t.x$\sec \left[\tan \left(x^4+4\right)\right]$
AnswerLet $y=\sec \left[\tan \left(x^4+4\right)\right]$Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left\{\sec \left[\tan \left(x^4+4\right)\right]\right\} \\ & =\sec \left[\tan \left(x^4+4\right)\right] \cdot \tan \left[\tan \left(x^4+4\right)\right] \cdot \frac{d}{d x}\left[\tan \left(x^4+4\right)\right] \\ & =\sec \left[\tan \left(x^4+4\right)\right] \cdot \tan \left[\tan \left(x^4+4\right)\right] \cdot \sec ^2\left(x^4+4\right) \cdot \frac{d}{d x}\left(x^4+4\right) \\ & =\sec \left[\tan \left(x^4+4\right)\right] \cdot \tan \left[\tan \left(x^4+4\right)\right] \cdot \sec ^2\left(x^4+4\right)\left(4 x^3+0\right) \\ & =4 x^3 \sec ^2\left(x^4+4\right) \cdot \sec \left[\tan \left(x^4+4\right)\right] \cdot \tan \left[\tan \left(x^4+4\right)\right]\end{aligned}$
View full question & answer→Question 882 Marks
Diffrentiate the following w.r.t.x$\tan [\cos (\sin x)]$
AnswerLet y = tan[cos (sinx)] Differentiating w.r.t. x, we get$\frac{d y}{d x}=\frac{d}{d x}\{\tan [\cos (\sin x)]\}$
$\begin{aligned} & =\sec ^2[\cos (\sin x)] \cdot \frac{d}{d x}[\cos (\sin x)] \\ & =\sec ^2[\cos (\sin x)] \cdot[-\sin (\sin x)] \cdot \frac{d}{d x}(\sin x) \\ & =-\sec ^2[\cos (\sin x)] \cdot \sin (\sin x) \cdot \cos x .\end{aligned}$
View full question & answer→Question 892 Marks
Diffrentiate the following w.r.t.x$\cos ^2\left[\log \left(x^2+7\right)\right]$
AnswerLet $y=\cos ^2\left[\log \left(x^2+7\right)\right]$Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left\{\cos \left[\log \left(x^2+7\right)\right]\right\}^2 \\ & =2 \cos \left[\log \left(x^2+7\right)\right] \cdot \frac{d}{d x}\left\{\cos \left[\log \left(x^2+7\right)\right]\right\} \\ & =2 \cos \left[\log \left(x^2+7\right)\right] \cdot\left\{-\sin \left[\log \left(x^2+7\right)\right]\right\} \cdot \frac{d}{d x}\left[\log \left(x^2+7\right)\right] \\ & =-2 \sin \left[\log \left(x^2+7\right)\right] \cdot \cos \left[\log \left(x^2+7\right)\right] \times \frac{1}{x^2+7} \cdot \frac{d}{d x}\left(x^2+7\right) \\ & =-\sin \left[2 \log \left(x^2+7\right)\right] \times \frac{1}{x^2+7} \cdot(2 x+0) \\ & =\frac{-2 x \cdot \sin \left[2 \log \left(x^2+7\right)\right]}{x^2+7} .\end{aligned}$
View full question & answer→Question 902 Marks
Diffrentiate the following w.r.t.x$e^{3 \sin 2 x-2 \cos 2 x}$
AnswerLet $y= e ^{3 \sin 2 x-2 \cos 2 x}$Differentiating w.r.t. x, we get
$\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[e^{3 \sin ^2 x-2 \cos ^2 x}\right] \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot \frac{d}{d x}\left(3 \sin ^2 x-2 \cos ^2 x\right) \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot\left[3 \frac{d}{d x}(\sin x)^2-2 \frac{d}{d x}(\cos x)^2\right] \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot\left[3 \times 2 \sin x \cdot \frac{d}{d x}(\sin x)-2 \times 2 \cos x \cdot \frac{d}{d x}(\cos x)\right] \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot[6 \sin x \cos x-4 \cos x(-\sin x)] \\ & =e^{3 \sin ^2 x-2 \cos ^2 x} \cdot(10 \sin x \cos x) \\ & =5\left(2 \sin x \cos ^2 x\right) \cdot e^{3 \sin 2 x-2 \cos ^2 x} \\ & =5 \sin 2 x \cdot e^{3 \sin ^2 x-2 \cos ^2 x} .\end{aligned}$
View full question & answer→Question 912 Marks
Diffrentiate the following w.r.t.x$\log \left[\cos \left(x^3-5\right)\right]$
AnswerLet $y=\log \left[\cos \left(x^3-5\right)\right]$Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left\{\log \left[\cos \left(x^3-5\right)\right]\right\}$
$\begin{aligned} & =\frac{1}{\cos \left(x^3-5\right)} \cdot \frac{d}{d x}\left[\cos \left(x^3-5\right)\right] \\ & =\frac{1}{\cos \left(x^3-5\right)} \cdot\left[-\sin \left(x^3-5\right)\right] \cdot \frac{d}{d x}\left(x^3-5\right) \\ & =-\tan \left(x^3-5\right) \times\left(3 x^2-0\right) \\ & =-3 x^2 \tan \left(x^3-5\right) .\end{aligned}$
View full question & answer→Question 922 Marks
Diffrentiate the following w.r.t.x$\operatorname{cosec}(\sqrt{\cos X})$
AnswerLet $y =\operatorname{cosec}(\sqrt{\cos X})$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}[\operatorname{cosec}(\sqrt{\cos x})] \\ & =-\operatorname{cosec}(\sqrt{\cos x}) \cdot \cot (\sqrt{\cos x}) \cdot \frac{d}{d x} \sqrt{\cos x} \\ & =-\operatorname{cosec}(\sqrt{\cos x}) \cdot \cot (\sqrt{\cos x}) \cdot \frac{1}{2 \sqrt{\cos x}} \cdot \frac{d}{d x}(\cos x) \\ & =-\operatorname{cosec}(\sqrt{\cos x}) \cdot \cot (\sqrt{\cos x}) \cdot \frac{1}{2 \sqrt{\cos x}} \cdot(-\sin x) \\ & =\frac{\sin x \cdot \operatorname{cosec}(\sqrt{\cos x}) \cdot \cot (\sqrt{\cos x})}{2 \sqrt{\cos x}} .\end{aligned}$
View full question & answer→Question 932 Marks
Diffrentiate the following w.r.t.x$5^{\sin 3 x+3}$
AnswerLet $y=5^{\sin 3 x+3}$Differentiating w.r.t. x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(5^{\sin ^3 x+3}\right)$
$\begin{aligned} & =5^{\sin ^3 x+3} \cdot \log 5 \cdot \frac{d}{d x}\left(\sin ^3 x+3\right) \\ & =5^{\sin ^3 x+3} \cdot \log 5 \cdot\left[3 \sin ^2 x \cdot \frac{d}{d x}(\sin x)+0\right] \\ & =5^{\sin ^3 x+3} \cdot \log 5 \cdot\left[3 \sin ^2 x \cos x\right] \\ & =3 \sin ^2 x \cos x \cdot 5^{\sin ^3 x+3} \cdot \log 5\end{aligned}$
View full question & answer→Question 942 Marks
Diffrentiate the following w.r.t.x$\cot ^3\left[\log \left(x^3\right)\right]$
AnswerLet $y=\cot ^3\left[\log \left(x^3\right)\right]$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[\cot \left(\log x^3\right)\right]^3 \\ & =3\left[\cot \left(\log x^3\right)\right]^2 \cdot \frac{d}{d x}\left[\cot \left(\log x^3\right)\right] \\ & =3 \cot ^2\left[\log \left(x^3\right)\right] \cdot\left[-\operatorname{cosec}^2\left(\log x^3\right)\right] \cdot \frac{d}{d x}\left(\log x^3\right) \\ & =-3 \cot ^2\left[\log \left(x^3\right)\right] \cdot \operatorname{cosec} 2\left[\log \left(x^3\right)\right] \cdot 3 \frac{d}{d x}(\log x) \\ & =-3 \cot ^2\left[\log \left(x^3\right)\right] \cdot \operatorname{cosec}^2\left[\log \left(x^3\right)\right] \cdot 3 \times \frac{1}{x} \\ & =\frac{-9 \operatorname{cosec}^2\left[\log \left(x^3\right)\right] \cdot \cot ^2\left[\log \left(x^3\right)\right]}{x} .\end{aligned}$
View full question & answer→Question 952 Marks
Diffrentiate the following w.r.t.x$\sqrt{\tan \sqrt{x}}$
AnswerLet $y =\sqrt{\tan \sqrt{x}}$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(\sqrt{\tan \sqrt{x}}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \cdot \frac{d}{d x}(\tan \sqrt{x}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \times \sec ^2 \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x}) \\ & =\frac{1}{2 \sqrt{\tan \sqrt{x}}} \times \sec ^2 \sqrt{x} \times \frac{1}{2 \sqrt{x}} \\ & =\frac{\sec ^2 \sqrt{x}}{4 \sqrt{x} \sqrt{\tan \sqrt{x}}} .\end{aligned}$
View full question & answer→Question 962 Marks
Diffrentiate the following w.r.t.x$\log \left[\tan \left(\frac{x}{2}\right)\right]$
AnswerLet $y=\log \left[\tan \left(\frac{x}{2}\right)\right]$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x} \log \left[\tan \left(\frac{x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{x}{2}\right)} \cdot \frac{d}{d x}\left[\tan \left(\frac{x}{2}\right)\right] \\ & =\frac{1}{\tan \left(\frac{x}{2}\right)} \cdot \sec ^2\left(\frac{x}{2}\right) \cdot \frac{d}{d x}\left(\frac{x}{2}\right)\end{aligned}$
$\begin{aligned} & =\frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)} \cdot \frac{1}{\cos ^2\left(\frac{x}{2}\right)} \cdot \frac{1}{2} \times 1 \\ & =\frac{1}{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)} \\ & =\frac{1}{\sin x}=\operatorname{cosec} x .\end{aligned}$
View full question & answer→Question 972 Marks
Diffrentiate the following w.r.t.x$\sqrt{e^{(3 x+2)}+5}$
AnswerLet $y =\sqrt{e^{(3 x+2)}+5}$Differentiating w.r.t. x, we get
$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[e^{(3 x+2)}+5\right]^{\frac{1}{2}} \\ & =\frac{1}{2}\left[e^{(3 x+2)}+5\right]^{-\frac{1}{2}} \cdot \frac{d}{d x}\left[e^{(3 x+2)}+5\right] \\ & =\frac{1}{2 \sqrt{e^{(3 x+2)}+5}} \cdot\left[e^{(3 x+2)} \cdot \frac{d}{d x}(3 x+2)+0\right] \\ & =\frac{1}{2 \sqrt{e^{(3 x+2)}+5}} \cdot\left[e^{(3 x+2)} \cdot(3 \times 1+0)\right] \\ & =\frac{3 e^{(3 x+2)}}{2 \sqrt{e^{(3 x+2)}+5}} \cdot\end{aligned}$
View full question & answer→Question 982 Marks
Diffrentiate the following w.r.t.x$\cos \left(x^2+a^2\right)$
AnswerLet $y=\cos \left(x^2+a^2\right)$ Differentiating w.r.t. $x$, we get $\begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}\left[\cos \left(x^2+a^2\right)\right] \\ & \left.=-\sin \left(x^2+a^2\right) \cdot \frac{d}{d x} x^2+a^2\right) \\ & =-\sin \left(x^2+a^2\right) \cdot(2 x+0) \\ & =-2 x \sin \left(x^2+a^2\right) \end{aligned}$
View full question & answer→