Question 12 Marks
Differentiate the function $\cos \left( {a\cos x + b\sin x} \right)$ w.r.t x for some constant a and b.
AnswerLet $y = \cos \left( {a\cos x + b\sin x} \right)$ for some constants a and b $\therefore \frac{{dy}}{{dx}} = - \sin \left( {a\cos x + b\sin x} \right)\frac{d}{{dx}}\left( {a\cos x + b\sin x} \right)$
$\Rightarrow \frac{{dy}}{{dx}} = - \sin \left( {a\cos x + b\sin x} \right)\left( { - a\sin x + b\cos x} \right)$
$\Rightarrow \frac{{dy}}{{dx}} = - \left( { - a\sin x + b\cos x} \right)\sin \left( {a\cos x + b\sin x} \right)$
$\Rightarrow \frac{{dy}}{{dx}} = \left( {a\sin x - b\cos x} \right)\sin \left( {a\cos x + b\sin x} \right)$
View full question & answer→Question 22 Marks
Differentiate the function $(5x)^{3\ \cos 2x}$ w.r.t to $x$.
AnswerLet $y = (5x)^{3\ \cos\ 2x}$
Then, $\log y = \log (5x)^{3 \cos 2x}$
$\Rightarrow \log y=3 \cos 2 x \times \log 5 x$
Differentiating both sides with respect to $x$, we get
$\frac{1}{y} \frac{d y}{d x}$ = $3\left[\log 5 x \times \frac{d}{d x}(\cos 2 x)+\cos 2 x \times \frac{d}{d x}(\log 5 x)\right]$ ...[$\because$ $\frac{d}{d x}(u v)=u \times \frac{d v}{d x}+v \times \frac{d u}{d x}$]
$\Rightarrow$ $\frac{\mathrm{dy}}{\mathrm{dx}}=3 \mathrm{y}\left[\log 5 \mathrm{x}(-2 \sin 2 \mathrm{x}) \times \frac{\mathrm{d}}{\mathrm{dx}}(2 \mathrm{x})+\cos 2 \mathrm{x} \times \frac{1}{5 \mathrm{x}} \times \frac{\mathrm{d}}{\mathrm{dx}}(5 \mathrm{x})\right]$
$\Rightarrow$ $\frac{d y}{d x}=3 y\left[-2 \sin 2 x \log 5 x+\frac{\cos 2 x}{x}\right]$
$\Rightarrow$ $\frac{d y}{d x}=y\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]$
$\Rightarrow$ $\frac{d y}{d x}=(5 x)^{3} \cos 2 x\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]$
$\therefore$ $\frac{d y}{d x}=(5 x)^{3} \cos 2 x\left[\frac{3 \cos 2 x}{x}-6 \sin 2 x \log 5 x\right]$
View full question & answer→Question 32 Marks
Differentiate the function $\sin^3x + \cos^6 x$, w.r.t to $x$.
AnswerLet $y = \sin^3 x + \cos^6 x$
Differentiating both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{3} x\right)+\frac{d}{d x}\left(\cos ^{6} x\right)$
$= 3 \sin ^{2} x \times \frac{d}{d x}(\sin x)+6 \cos ^{5} x \times \frac{d}{d x}(\cos x)$
$=3 \sin ^{2} x \times \cos x+6 \cos ^{5} x \times(-\sin x)$ $[ \because~\left.\frac{d}{d x}(\sin x)=\cos x\ \& \frac{d}{d x}(\cos x)=-\sin x\right]$
$= 3 \sin x \cos x\left(\sin x-2 \cos ^{4} x\right)$
$\therefore$ $\frac{d y}{d x}=3 \sin x \cos x\left(\sin x-2 \cos ^{4} x\right)$
View full question & answer→Question 42 Marks
Differentiate the function ${\left( {3{x^2} - 9x + 5} \right)^9}$ w.r.t to x.
AnswerLet $y = {\left( {3{x^2} - 9x + 5} \right)^9}$ $\therefore \frac{{dy}}{{dx}} = 9{\left( {3{x^2} - 9x + 5} \right)^8}\frac{d}{{dx}}\left( {3{x^2} - 9x + 5} \right)$
$\left[ {\because \frac{d}{{dx}}{{\left\{ {f\left( x \right)} \right\}}^4} = n{{\left\{ {f\left( x \right)} \right\}}^{n - 1}}\frac{d}{{dx}}f\left( x \right)} \right]$
$\Rightarrow \frac{{dy}}{{dx}} = 9{\left( {3{x^2} - 9x + 5} \right)^8}\left[ {3\left( {2x} \right) - 9\left( 1 \right) + 0} \right] = 9{\left( {3{x^2} - 9x + 5} \right)^8}\left[ {6x - 9} \right]$
$ \Rightarrow \frac{{dy}}{{dx}} = 27{\left( {3{x^2} - 9x + 5} \right)^8}\left[ {2x - 3} \right]$
View full question & answer→Question 52 Marks
Find the second-order derivative of the function log x
AnswerLet y = log x
Now,
$\frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x}$
Therefore,
$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{1}{x}\right)=\left(-\frac{1}{x^{2}}\right)$
View full question & answer→Question 62 Marks
Find the second-order derivative of the function x.cos x
AnswerLet y = x.cos x
$\therefore \frac{{dy}}{{dx}} = x\frac{d}{{dx}}\cos x + \cos x\frac{d}{{dx}}x = - x\sin x + \cos x$
$\Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = - \frac{d}{{dx}}\left( {x\sin x} \right) + \frac{d}{{dx}}\cos x$
$= - \left[ {x\frac{d}{{dx}}\sin x + \sin x\frac{d}{{dx}}x} \right] - \sin x$
= -(x cos x + sin x) - sin x
= - x cos x - sin x - sin x
= -x cos x - 2 sin x
= -(x cos x + 2 sin x).
Which is the required solution.
View full question & answer→Question 72 Marks
Find the second-order derivatives of the function $x^{20}$
AnswerLet us take $y = x^{20}$
Now,
$\frac{d y}{d x}=\frac{d}{d x}(x^{20})$
$= 20 x^{19}$
Therefore,
$\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(20 x^{19}\right)=20 \frac{d}{d x}{\left(x^{19}\right)}$ = $20 \times 19 \times x^{18}$
$= 380 x^{18}$
View full question & answer→Question 82 Marks
Find the second-order derivatives of the function $x^2 + 3x + 2$
AnswerLet us take $y = x^2 + 3x + 2$
Now,
$\frac{d y}{d x}=\frac{d\left(x^{2}\right)}{d x}+\frac{d(3 x)}{d x}+\frac{d(2)}{d x}$
$= 2x + 3$
Therefore,
$\frac{d^{2} y}{d x^{2}}=\frac{d(2 x+3)}{d x}=\frac{d(2 x)}{d x}+\frac{d(3)}{d x}$
$= 2 + 0$
$= 2$
View full question & answer→Question 92 Marks
If x and y are connected parametrically by the equation $x = a\sec \theta,$ $y = b\tan \theta $, without eliminating the parameter, find $\frac{{dy}}{{dx}}$.
AnswerGiven: $x = a\sec \theta$ and $y = b\tan \theta $
$\therefore \frac{{dx}}{{d\theta }} = a\sec \theta \tan \theta$ and $\frac{{dy}}{{d\theta }} = b{\sec ^2}\theta$
$\therefore \frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{b{{\sec }^2}\theta }}{{a\sec \theta \tan \theta }}$
$= \frac{{b\sec \theta }}{{a\tan \theta }}$
$= \frac{{b.\frac{1}{{\cos \theta }}}}{{a.\frac{{\sin \theta }}{{\cos \theta }}}}$
$= \frac{b}{{a\sin \theta }}$
$= \frac{b}{a}\cos ec\theta$
View full question & answer→Question 102 Marks
If x and y are connected parametrically by the equation x = a ($\theta$ – sin $\theta$), y = a (1 + cos $\theta$) without eliminating the parameter. Find $\frac{d y}{d x}$.
AnswerGiven functions are x = $a(\theta - sin \theta)$ and y = a(1 + cos$\theta$)
$\frac{d x}{d \theta}=a \frac{d}{d \theta}(\theta-\sin \theta)$
$\frac{d x}{d \theta}=a\left[\frac{d}{d \theta} \theta-\frac{d}{d \theta} \sin \theta\right]$
$\frac{d x}{d \theta}=a(1-\cos \theta)$
$\frac{d y}{d \theta}=a \frac{d}{d \theta}(1+\cos \theta)$
$\frac{d y}{d \theta}=a\left[\frac{d}{d \theta}(1)+\frac{d}{d \theta} \cos \theta\right]$
$\frac{d \mathrm{y}}{d \theta}=a(0-\sin \theta)$
= -a sin$\theta$
= $\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}$ = $\frac{-a \sin \theta}{a(1-\cos \theta)}=\frac{-\sin \theta}{1-\cos \theta}$
= $-\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin ^{2} \frac{\theta}{2}}$
= $-\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}$
= $-\cot \frac{\theta}{2}$
View full question & answer→Question 112 Marks
If x and y are connected parametrically by the equation $x = \cos \theta - \cos 2\theta ,y = \sin \theta - \sin 2\theta $, without eliminating the parameter, find $\frac{{dy}}{{dx}}.$
AnswerGiven: $x = \cos \theta - \cos 2\theta $ and $y = \sin \theta - \sin 2\theta$ $\therefore \frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}\cos \theta - \frac{d}{{d\theta }}\cos 2\theta $ and $\frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\sin \theta - \frac{d}{{d\theta }}\sin 2\theta$
$\Rightarrow \frac{{dx}}{{d\theta }} = - \sin \theta - \left( { - \sin 2\theta } \right)\frac{d}{{d\theta }}2\theta$ and $\frac{{dy}}{{d\theta }} = \cos \theta - \cos 2\theta \frac{d}{{d\theta }}2\theta$
$\Rightarrow \frac{{dx}}{{d\theta }} = - \sin \theta + \left( {\sin 2\theta } \right)2$ and $\frac{{dy}}{{d\theta }} = \cos \theta - \cos 2\theta \times 2$
$\Rightarrow \frac{{dx}}{{d\theta }} = 2\sin 2\theta - \sin \theta $ and $\frac{{dy}}{{d\theta }} = \cos \theta - 2\cos 2\theta $
Now $\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{\cos \theta - 2\cos 2\theta }}{{2\sin 2\theta - \sin \theta }}$
View full question & answer→Question 122 Marks
If x and y are connected parametrically by the equation x = 4t, $y = \frac{4}{t}$, without eliminating the parameter, find $\frac{{dy}}{{dx}}.$
AnswerGiven: x = 4t and $y = \frac{4}{t}$ $\therefore \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {4t} \right) = 4\frac{d}{{dt}}t = 4$
and $\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {\frac{4}{t}} \right) = \frac{{t\frac{d}{{dt}}4 - 4\frac{d}{{dt}}t}}{{{t^2}}}$
$\Rightarrow \frac{{dy}}{{dt}} = \frac{{t \times 0 - 4 \times 1}}{{{t^2}}} = - \frac{4}{{{t^2}}}$
Now $\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{ - \frac{4}{{{t^2}}}}}{4} = \frac{{ - 1}}{{{t^2}}}$
View full question & answer→Question 132 Marks
If x and y are connected parametrically by the equation x = sin t, y = cos 2t, without eliminating the parameter, find $\frac{{dy}}{{dx}}.$
AnswerGiven: x = sin t and y = cos 2t $\therefore \frac{{dx}}{{dt}} = \cos t$ and $\frac{{dy}}{{dt}} = - \sin 2t\frac{d}{{dt}}\left( {2t} \right) = - 2\sin 2t$
Now $\frac{{dy}}{{dt}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{ - 2\sin 2t}}{{\cos t}} $ $= \frac{{ - 2 \times 2\sin t\cos t}}{{\cos t}} = - 4\sin t$
View full question & answer→Question 142 Marks
If x and y are connected parametrically by the equation $x = a\cos \theta ,y = b\cos \theta$, without eliminating the parameter, Find $\frac{{dy}}{{dx}}.$
AnswerGiven: $x = a\cos \theta$ and $y = b\cos \theta$ $\therefore \frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}\left( {a\cos \theta } \right)$and $\frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left( {b\cos \theta } \right)$
$\Rightarrow \frac{{dx}}{{d\theta }} = a\frac{d}{{d\theta }}\left( {\cos \theta } \right)$ and $\frac{{dy}}{{d\theta }} = b\frac{d}{{d\theta }}\left( {\cos \theta } \right)$
$ \Rightarrow \frac{{dx}}{{d\theta }} = - a\sin \theta $ and $\frac{{dy}}{{d\theta }} = - b\sin \theta$
Now $\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = - \frac{{ - b\sin \theta }}{{ - a\sin \theta }} = \frac{b}{a}$
View full question & answer→Question 152 Marks
If x and y are connected parametrically by the equation $x = 2at^2, y = at^4,$ without eliminating the parameter, Find$\frac{{dy}}{{dx}}.$
AnswerGiven: $x = 2at^2$ and $y = at^4 \therefore \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left( {2a{t^2}} \right)$ and $\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {a{t^4}} \right)$
$ \Rightarrow \frac{{dx}}{{dt}} = 2a\frac{d}{{dt}}\left( {{t^2}} \right) = 2a.2t = 4at$ and $\frac{{dy}}{{dt}} = a\frac{d}{{dt}}\left( {{t^4}} \right) = a.4{t^3} = 4a{t^3}$
Now $\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{4a{t^3}}}{{4at}} = {t^2}$
View full question & answer→Question 162 Marks
Differentiate the function ${\left( {\log x} \right)^{\cos x}}$ w.r.t. x.
AnswerLet $y = {\left( {\log x} \right)^{\cos x}}$ ……….(i) Taking log on both sides, we have
$\Rightarrow \log y = \log {\left( {\log x} \right)^{\cos x}} = \cos x\log \left( {\log x} \right)$
$\Rightarrow \frac{d}{{dx}}\log y = \frac{d}{{dx}}\left[ {\cos x\log \left( {\log x} \right)} \right]$
$\Rightarrow \frac{1}{y}\frac{dy}{{dx}} = \cos x\frac{d}{{dx}}\log \left( {\log x} \right) + \log \left( {\log x} \right)\frac{d}{{dx}}\cos x$ [By Product rule]
$ \Rightarrow \frac{1}{y}.\frac{{dy}}{{dx}} = \cos x\frac{1}{{\log x}}\frac{d}{{dx}}\left( {\log x} \right) + \log \left( {\log x} \right)\left( { - \sin x} \right)$
$\Rightarrow \frac{1}{y}.\frac{{dy}}{{dx}} = \frac{{\cos x}}{{\log x}}.\frac{1}{x} - \sin x\log \left( {\log x} \right)$
$\Rightarrow \frac{{dy}}{{dx}} = y\left[ {\frac{{\cos x}}{{x\log x}} - \sin x\log \left( {\log x} \right)} \right]$
$\Rightarrow \frac{{dy}}{{dx}} = {\left( {\log x} \right)^{\cos x}}\left[ {\frac{{\cos x}}{{x\log x}} - \sin x\log \left( {\log x} \right)} \right]$
View full question & answer→Question 172 Marks
If u, v and w are functions of x, then show that
$\frac{d}{d x}(u . v . w)=\frac{d u}{d x} v . w+u . \frac{d v}{d x} \cdot w+u \cdot v \frac{d w}{d x}$
in two ways - first by repeated application of product rule, second by logarithmic differentiation.
AnswerTo prove: $\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{u} . \mathrm{v} . \mathrm{w})=\frac{\mathrm{du}}{\mathrm{dx}} \mathrm{v} \cdot \mathrm{w}+\mathrm{u} \cdot \frac{\mathrm{dv}}{\mathrm{dx}} \cdot \mathrm{w}+\mathrm{u.v.} \frac{\mathrm{dw}}{\mathrm{dx}}$
Let y = u.v.w = u.(v.w)
- By applying product rule differentiate both sides with respect to x
$\frac{d y}{d x}=(v \cdot w) \cdot \frac{d u}{d x}+u \cdot \frac{d}{d x}(v \cdot w)$
$\Rightarrow \frac{d y}{d x}=(v \cdot w) \cdot \frac{d u}{d x}+u \cdot\left[v \cdot \frac{d}{d x}(w)+w \cdot \frac{d}{d x}(v)\right]$
$\Rightarrow \frac{d y}{d x}=(v \cdot w) \cdot \frac{d u}{d x}+(u \cdot v) \cdot \frac{d w}{d x}+(u . w) \cdot \frac{d v}{d x}$ - Taking log on both sides, we get
as, y = u.v.w
log y = log (u.v.w)
log y = log u + log v + log w
Now, differentiate both sides with respect to x
$\Rightarrow \frac{d}{d x}(\log y)=\frac{d}{d x} \log u+\frac{d}{d x} \log v+\frac{d}{d x} \log w$
$\Rightarrow \frac{1}{\mathrm{y}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{y})=\frac{1}{\mathrm{u}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{u})+\frac{1}{\mathrm{v}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{v})+\frac{1}{\mathrm{w}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{w})$
$\Rightarrow \frac{d}{d x}(y)=y\left[\frac{1}{u} \cdot \frac{d u}{d x}+\frac{1}{v} \cdot \frac{d v}{d x}+\frac{1}{w} \cdot \frac{d w}{d x}\right]$
$\Rightarrow \frac{d y}{d x}=u . v . w\left[\frac{1}{u} \cdot \frac{d u}{d x}+\frac{1}{v} \cdot \frac{d v}{d x}+\frac{1}{w} \cdot \frac{d w}{d x}\right]$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v} \cdot \mathrm{w} \cdot \frac{\mathrm{du}}{\mathrm{dx}}+\mathrm{u.w} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}+\mathrm{u.v.} \frac{\mathrm{dw}}{\mathrm{dx}}$
View full question & answer→Question 182 Marks
Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
- by using product rule
- by expanding the product to obtain a single polynomial
- by logarithmic differentiation.
Do they all give the same answer?
Answer
- Given: $(x^2 – 5x + 8) (x^3 + 7x + 9)$
Let y $= (x^2 – 5x + 8) (x^3 + 7x + 9)$
By applying product rule differentiate both sides with respect to $x$
$\frac{d y}{d x}=\frac{d}{d x} (x^2 − 5x + 8)(x^3 + 7x + 9)$
$\Rightarrow \frac{d y}{d x}=\left(x^{3}+7 x+9\right) \cdot \frac{d}{d x}\left(x^{2}-5 x+8\right)$ + $\left(x^{2}-5 x+8\right) \cdot \frac{d}{d x}\left(x^{3}+7 x+9\right)$
$\Rightarrow \frac{d y}{d x}=\left(x^{3}+7 x+9\right) \cdot(2 x-5)+\left(x^{2}-5 x+8\right) \cdot\left(3 x^{2}+7\right)$ $\Rightarrow \frac{{dy}}{{dx}}=2{x}^{4} +14 {x}^{2}+18 \mathrm{x}-5{x}^{3} - 35 x − 45 + 3 x ^4+ 7 x ^2− 15 x ^3− 35 x + 24 x ^2 + 56$
$\Rightarrow \frac{d y}{d x}= 5x^4− 20x^3+ 45x^2− 52x + 11$
- Given:$ (x^2 – 5x + 8) (x^3 + 7x + 9)$
Let $y = (x^2 – 5x + 8) (x^3 + 7x + 9)$
$\Rightarrow y = (x^2 – 5x + 8) (x^3 + 7x + 9)$
$\Rightarrow y = x^5 + 7x^3 + 9x^2 - 5x^4 – 35x^2 - 45x + 8x^3 + 56x + 72$
$\Rightarrow y = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72$
Now, differentiate both sides with respect to x, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(x^{5}\right)-\frac{d}{d x}\left(5 x^{4}\right)+\frac{d}{d x}\left(15 x^{3}\right)-\frac{d}{d x}\left(26 x^{2}\right)+\frac{d}{d x}(11 x)+\frac{d}{d x}(72)$
$\frac{d y}{d x} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$
- Given: $(x^2 – 5x + 8) (x^3 + 7x + 9)$
Let $y = (x^2 – 5x + 8) (x^3 + 7x + 9)$
Taking log on both sides, we get
$\log y = \log ((x^2 – 5x + 8) (x^3 + 7x + 9))$
$\Rightarrow \log y = \log (x^2 – 5x + 8) + \log (x^3 + 7x + 9)$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})=\frac{\mathrm{d}}{\mathrm{dx}} \log \left(\mathrm{x}^{2}-5 \mathrm{x}+8\right)+\frac{\mathrm{d}}{\mathrm{dx}} \log \left(\mathrm{x}^{3}+7 \mathrm{x}+9\right)$
$\Rightarrow \frac{1}{\mathrm{y}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{y})=\left[\frac{1}{\left(\mathrm{x}^{2}-5 \mathrm{x}+8\right)} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-5 \mathrm{x}+8\right)+\frac{1}{\left(\mathrm{x}^{3}+7 \mathrm{x}+9\right)} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}+7 \mathrm{x}+9\right)\right]$
$\Rightarrow \frac{1}{y} \frac{d}{d x}(y)=\left[\frac{1}{\left(x^{2}-5 x+8\right)} \cdot(2 x-5)+\frac{1}{\left(x^{3}+7 x+9\right)} \cdot\left(3 x^{2}+7\right)\right]$
$\Rightarrow \frac{d}{d x}(y)=y \cdot\left[\frac{(2 x-5)}{\left(x^{2}-5 x+8\right)}+\frac{\left(3 x^{2}+7\right)}{\left(x^{3}+7 x+9\right)}\right]$
$\Rightarrow \frac{d}{d x}(y)=y \cdot\left[\frac{(2 x-5)\left(x^{3}+7 x+9\right)+\left(3 x^{2}+7\right)\left(x^{2}-5 x+8\right)}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]$
$\Rightarrow \frac{d}{d x}(y)$$=y \cdot\left[\frac{2 x^{4}+14 x^{2}+18 x-5 x^{3}-35 x-45+3 x^{4}-15 x^{3}+24 x^{2}+7 x^{2}-35 x+56}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]$
$\Rightarrow \frac{d}{d x}(y)=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$.$\left[\frac{5 x^{4}-20 x^{3}-45 x^{2}-52 x+11}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]$
$\Rightarrow \frac{d y}{d x} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$
View full question & answer→Question 192 Marks
Find the derivative of the function given by $f(x) = (1 + x) (1 + x^2) (1 + x^4) (1 + x^8)$ and hence find f ′(1).
AnswerGiven: $f(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)$
Taking log on both sides, we get
log $f(x) = \log (1 + x) + \log(1 + x^2) + \log (1 + x^4) + \log (1 + x^8)$
Now, differentiate both sides with respect to x
$\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{f}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}} \log (1+\mathrm{x})+\frac{\mathrm{d}}{\mathrm{dx}} \log \left(1+\mathrm{x}^{2}\right)+\frac{\mathrm{d}}{\mathrm{dx}} \log \left(1+\mathrm{x}^{4}\right)+\frac{\mathrm{d}}{\mathrm{dx}} \log \left(1+\mathrm{x}^{8}\right)$
$\Rightarrow \frac{1}{f(x)} \cdot \frac{d}{d x}[f(x)]$ = $\frac{1}{1+\mathrm{x}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(1+\mathrm{x})+\frac{1}{1+\mathrm{x}^{2}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(1+\mathrm{x}^{2}\right)+\frac{1}{1+\mathrm{x}^{4}} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(1+\mathrm{x}^{4}\right)$ + $\frac{1}{1+x^{8}} \cdot \frac{d}{d x}\left(1+x^{8}\right)$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x})\left[\frac{1}{1+\mathrm{x}}+\frac{1}{1+\mathrm{x}^{2}} \cdot(2 \mathrm{x})+\frac{1}{1+\mathrm{x}^{4}} \cdot\left(4 \mathrm{x}^{3}\right)+\frac{1}{1+\mathrm{x}^{8}} \cdot\left(8 \mathrm{x}^{7}\right)\right]$
$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=(1+\mathrm{x})\left(1+\mathrm{x}^{2}\right)\left(1+\mathrm{x}^{4}\right)\left(1+\mathrm{x}^{8}\right)\left[\frac{1}{1+\mathrm{x}}+\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}+\frac{4 \mathrm{x}^{3}}{1+\mathrm{x}^{4}}+\frac{8 \mathrm{x}^{7}}{1+\mathrm{x}^{8}}\right]$
$\Rightarrow \mathrm{f}^{\prime}(1)=(1+1)\left(1+1^{2}\right)\left(1+1^{4}\right)(1+1^8)$ $\left[\frac{1}{1+1}+\frac{2(1)}{1+1}+\frac{4(1)^{3}}{1+(1)^{4}}+\frac{8(1)^{7}}{1+(1)^{8}}\right]$
$\Rightarrow \mathrm{f}^{\prime}(1)=(2)(2)(2)(2)\left[\frac{1}{2}+\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\right]$
$\Rightarrow \mathrm{f}^{\prime}(1)=16\left(\frac{1+2+4+8}{2}\right)$
$\Rightarrow \mathrm{f}^{\prime}(1)=16\left(\frac{15}{2}\right)$
$\Rightarrow \mathrm{f}^{\prime}(1)=120$
View full question & answer→Question 202 Marks
Find $\frac{d y}{d x}$ of the function $xy = e^{(x – y)}$
AnswerGiven: $xy = e^{(x – y)}$
Taking log on both sides, we get
$\log (x y) = \log (e^{(x – y)})$
$\Rightarrow \log x + \log y = (x - y) \log e$
$\Rightarrow \log x + \log y = (x - y) .1$
$\Rightarrow \log x + \log y = (x - y)$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}+\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{y}=\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}-\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{y}$
$\implies$ $\frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}$
$\implies$$\left(1+\frac{1}{y}\right) \frac{d y}{d x}=1-\frac{1}{x}$
$\implies$$\frac{1+y}{y} \frac{d y}{d x}=\frac{x-1}{x}$
$\implies$$\frac{d y}{d x}=\frac{y(x-1)}{x(1+y)}$
View full question & answer→Question 212 Marks
Find $\frac { d y } { d x }$ of the function $(\cos x)^y = (\cos y)^x.$
AnswerWe have, $(\cos x)^y = (\cos y)^x$ On taking log both sides, we get
$\log(\cos x)^y = \log(\cos y)^x$
$\Rightarrow y \log(\cos x) = x \log(\cos y)$
On differentiating both sides w.r.t x, we get
$y \cdot \frac { d } { d x } \log ( \cos x ) + \log \cos x \cdot \frac { d } { d x } ( y )$
$= x \frac { d } { d x } \log \left( \cos y) + \log ( \cos y ) \frac { d } { d x } ( x )\right.$ [by using product rule of derivative]
$\Rightarrow \quad y \cdot \frac { 1 } { \cos x } \frac { d } { d x } ( \cos x ) + \log ( \cos x ) \frac { d y } { d x }$$= x \cdot \frac { 1 } { \cos y } \frac { d } { d x }$(cos y) + log cos y.1
$\Rightarrow y \cdot \frac { 1 } { \cos x } ( - \sin x ) + \log ( \cos x ) \cdot \frac { d y } { d x }$$= x \cdot \frac { 1 } { \cos y } $(-sin y)$\frac{dy}{dx}$ + log cos y.1
$\Rightarrow$ - y tanx + log(cos x)$\frac { d y } { d x }$ =-x tan y$\frac { d y } { d x }$+ log(cos y)
$\Rightarrow$[ x tan y + log (cos x)]$\frac{dy}{dx}$= log(cos y) + y tan x
$\therefore \quad \frac { d y } { d x } = \frac { \log ( \cos y ) + y \tan x } { x \tan y + \log ( \cos x ) }$
View full question & answer→Question 222 Marks
Find $\frac{{dy}}{{dx}}$, of the function $y^x = x^y$
AnswerGiven: $y^x = x^y$
$\Rightarrow {x^y} = {y^x}$
$\Rightarrow \log {x^y} = \log {y^x}$
$ \Rightarrow y\log x = x\log y$
$\Rightarrow \frac{d}{{dx}}\left( {y\log x} \right) = \frac{d}{{dx}}\left( {x\log y} \right)$
$\Rightarrow y.\frac{1}{x} + \log x.\frac{{dy}}{{dx}} = x.\frac{1}{y}\frac{{dy}}{{dx}} + \log y.1$
$\Rightarrow \left( {\log x - \frac{x}{y}} \right)\frac{{dy}}{{dx}} = \log y - \frac{y}{x}$
$\Rightarrow \left( {\frac{{y\log x - x}}{y}} \right)\frac{{dy}}{{dx}} = \frac{{x\log y - y}}{x}$
$\Rightarrow \frac{{dy}}{{dx}} = \frac{{y\left( {x\log y - y} \right)}}{{x\left( {y\log x - x} \right)}}$
View full question & answer→Question 232 Marks
Find $\frac{dy}{dx}$ of the function $x^y + y^x = 1$
AnswerGiven: $x^y + y^x = 1$
Let $y = x^y + y^x = 1$
Let $u = x^y$ and $v = y^x$
Then, $u + v = 1$
$\Rightarrow \frac{d u}{d x}+\frac{d v}{d x}=0$
For, $u = x^y$
Taking log on both sides, we get
$\log u = \log x^y$
$\Rightarrow \log u=y \cdot \log (x)$
Now, differentiating both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{u})=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{y} \cdot \log (\mathrm{x})]$
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\left\{\mathrm{y} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})+\log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{y})\right\}$
$\Rightarrow \frac{d u}{d x}=u\left[y \cdot \frac{1}{x}+\log x \cdot\left(\frac{d y}{d x}\right)\right]$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x}^{\mathrm{y}}\left[\frac{\mathrm{y}}{\mathrm{x}}+\log \mathrm{x} \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right]$
For $v = y^x$
Taking log on both sides, we get
$\log v = \log y^x$
$\Rightarrow \log v=x \cdot \log (y)$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{v})=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{x} \cdot \log (\mathrm{y})]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dx}}=\left\{\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})+\log \mathrm{y} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}\right\}$
$\Rightarrow \frac{d v}{d x}=v\left[x \cdot \frac{1}{y} \cdot \frac{d y}{d x}+\log y \cdot\left(\frac{d x}{d x}\right)\right]$
$\Rightarrow \frac{d v}{d x}=y^{x}\left[\frac{x}{y} \cdot \frac{d y}{d x}+\log y\right]$
because, $\frac{d u}{d x}+\frac{d v}{d x}=0$
So, $\mathrm{x}^{\mathrm{y}}\left[\frac{\mathrm{y}}{\mathrm{x}}+\log \mathrm{x} \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right]+\mathrm{y}^{\mathrm{x}}\left[\frac{\mathrm{x}}{\mathrm{y}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\log \mathrm{y}\right]=0$
$\Rightarrow\left(\mathrm{x}^{\mathrm{y}} \log \mathrm{x}+\mathrm{xy}^{\mathrm{x}-1}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\left(\mathrm{yx}^{\mathrm{y}-1}+\mathrm{y}^{\mathrm{x}} \log \mathrm{y}\right)=0$
$\Rightarrow\left(\mathrm{x}^{\mathrm{y}} \log \mathrm{x}+\mathrm{xy}^{\mathrm{x}-1}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=-\left(\mathrm{yx}^{\mathrm{y}-1}+\mathrm{y}^{\mathrm{x}} \log \mathrm{y}\right)$
$\frac{d y}{d x}=-\frac{\left(y x^{y-1}+y^{x} \log y\right)}{\left(x^{y} \log x+x y^{x-1}\right)}$
View full question & answer→Question 242 Marks
Differentiate the function ${\left( {x\cos x} \right)^x} + {\left( {x\sin x} \right)^{\frac{1}{x}}}$ w.r.t. x.
AnswerLet $y = {\left( {x\cos x} \right)^x} + {\left( {x\sin x} \right)^{\frac{1}{x}}}$Putting $u = (x \cos x)^x$ and $v = {\left( {x\sin x} \right)^{\frac{1}{x}}}$
we have y = u + v
$\therefore \frac{{dy}}{{dx}} = \frac{{du}}{{dx}} + \frac{{dv}}{{dx}}$ ....(i)
Now $u = (x \cos x)^x$
$\Rightarrow \log u = \log {\left( {x\cos x} \right)^x} = x\log \left( {x\cos x} \right)$
$\Rightarrow \log u = x\left( {\log x + \log \cos x} \right)$
$\Rightarrow \frac{d}{{dx}}\log u = \frac{d}{{dx}}\left\{ {x\left( {\log x + \log \cos x} \right)} \right\}$
$\Rightarrow \frac{1}{u}\frac{{du}}{{dx}}$ $ = x\left[ {\frac{1}{x} + \frac{1}{{\cos x}}.\left( { - \sin x} \right)} \right] + \left( {\log x + \log \cos x} \right).1$
$\Rightarrow \frac{1}{u}\frac{{du}}{{dx}} $$ = \left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right]$
$\Rightarrow \frac{{du}}{{dx}} = u\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right]$
$\Rightarrow \frac{{du}}{{dx}} = {\left( {x\cos x} \right)^x}\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right]$ .....(ii)
Again $v = {\left( {x\sin x} \right)^{\frac{1}{x}}}$
$\Rightarrow \log v = \log {\left( {x\sin x} \right)^{\frac{1}{x}}} = \frac{1}{x}\log \left( {x\sin x} \right)$
$\Rightarrow \log v = \frac{1}{x}\left( {\log x + \log \sin x} \right)$
$\Rightarrow \frac{d}{{dx}}\log v = \frac{d}{{dx}}\left\{ {\frac{1}{x}\left( {\log x + \log \sin x} \right)} \right\}$
$\Rightarrow \frac{1}{v}\frac{dv}{{dx}} $$= \frac{1}{x}\left[ {\frac{1}{x} + \frac{1}{{\ sinx }}.\cos x} \right] + \left( {\log x + \log \sin x} \right)\left( {\frac{{ - 1}}{{{x^2}}}} \right)$
$\Rightarrow \frac{1}{v}\frac{dv}{{dx}} $$= \left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$
$\Rightarrow \frac{dv}{{dx}} = v\left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$
$\Rightarrow \frac{dv}{{dx}} = {\left( {x\sin x} \right)^{\frac{1}{x}}}\left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$ ...(iii)
Putting the values from eq. (ii) and (iii) in eq. (i)
$\frac{d}{{dx}} = {\left( {x\cos x} \right)^{^x}}$$\left[ {1 - x\tan x + \log \left( {x\cos x} \right)} \right] + {\left( {x\sin x} \right)^{\frac{1}{x}}}$$\left[ {\frac{1}{{{x^2}}} + \frac{{\cot x}}{x} - \frac{{\log \left( {x\sin x} \right)}}{{{x^2}}}} \right]$
View full question & answer→Question 252 Marks
Differentiate the $\frac{{\cos x}}{{\log x}},x > 0$ w.r.t. x.
AnswerLet $y = \frac{{\cos x}}{{\log x}}$ $\therefore \frac{{dy}}{{dx}} = \frac{{\log x\frac{d}{{dx}}\left( {\cos x} \right) - \cos x\frac{d}{{dx}}\left( {\log x} \right)}}{{{{\left( {\log x} \right)}^2}}}$ [By quotient rule]
$= \frac{{\log x\left( { - \sin x} \right) - \cos x.\frac{1}{x}}}{{{{\left( {\log x} \right)}^2}}}$
$= \frac{{ - \left( {\sin x\log x + \frac{{\cos x}}{x}} \right)}}{{{{\left( {\log x} \right)}^2}}}$
$= \frac{{ - \left( {x\sin x\log x + \cos x} \right)}}{{x{{\left( {\log x} \right)}^2}}}$
View full question & answer→Question 262 Marks
Differentiate the log (log x), x > 1 w.r.t. x.
AnswerLet y = log(log x)
So, by using chain rule, we get
$\frac{d y}{d x}=\frac{d}{d x}(\log (\log x))$
$=\frac{1}{\log x} \cdot \frac{d}{d x}(\log x)$
$=\frac{1}{\log x} \cdot \frac{1}{x}$
$=\frac{1}{\mathrm{xlogx}}$
View full question & answer→Question 272 Marks
Differentiate the $\sqrt{e^{\sqrt{x}}}, x>0$ w.r.t. x.
AnswerLet $y=\sqrt{e^{\sqrt{x}}}$
Then, $y^{2}=e^{\sqrt{x}}$
Now, differentiating both sides w.r.t x, we get,
$2 y \frac{d y}{d x}=e^{\sqrt{x}} \frac{d}{d x}(\sqrt{x})$
$=e^{\sqrt{x}} \frac{1}{2} \cdot \frac{1}{\sqrt{x}}$
$\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 y \sqrt{x}}$
$\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{e^{\sqrt{x}}} \sqrt{x}}$
$\Rightarrow \frac{d y}{d x}=\frac{e^{\sqrt{x}}}{4 \sqrt{x e^{\sqrt{x}}}}$
View full question & answer→Question 282 Marks
Differentiate the ${e^x} + {e^{{x^2}}} + ... + {e^{{x^5}}}$ w.r.t. x.
AnswerLet $y = {e^x} + {e^{{x^2}}} + ... + {e^{{x^5}}} = {e^x} + {e^{{x^2}}} + {e^{{x^3}}} + {e^{{x^4}}} + {e^{{x^5}}}$ $\therefore \frac{{dy}}{{dx}} = \frac{d}{{dx}}{e^x} + \frac{d}{{dx}}{e^{{x^2}}} + \frac{d}{{dx}}{e^{{x^3}}} + \frac{d}{{dx}}{e^{{x^4}}} + \frac{d}{{dx}}{e^{{x^5}}}$
$= {e^x} + {e^{{x^2}}}\frac{d}{{dx}}{x^2} + {e^{{x^3}}}\frac{d}{{dx}}{x^3} + {e^{{x^4}}} \frac{d}{dx}{x^4}+ {e^{{x^5}}}\frac{d}{{dx}}{x^5}$
$= {e^x} + {e^{{x^2}}}.2x + {e^{{x^3}}}.3{x^2} + {e^{{x^4}}}.4{x^3} + {e^{{x^5}}}.5{x^4}$
$ = {e^x} + 2x.{e^{{x^2}}} + 3{x^2}{e^{{x^3}}} + 4{x^3}.{e^{{x^4}}} + 5{x^4}.{e^{{x^5}}}$
View full question & answer→Question 292 Marks
Differentiate the log$(\cos e^x)$ w.r.t. x.
AnswerLet $y = \log(\cos e^x) \therefore \frac{{dy}}{{dx}} = \frac{1}{{\cos {e^x}}}\frac{d}{{dx}}\left( {\cos {e^x}} \right)\,\,\left[ {\because \frac{d}{{dx}}\log f\left( x \right) = \frac{1}{{f\left( x \right)}}\frac{d}{{dx}}f\left( x \right)} \right]$
$= \frac{1}{{\cos {e^x}}}\left( { - \sin {e^x}} \right)\frac{d}{{dx}}{e^x}\,\,\left[ {\because \frac{d}{{dx}}\cos f\left( x \right) = - \sin f\left( x \right)\frac{d}{{dx}}f\left( x \right)} \right]$
$= - (tane^x)e^x = -e^x (\tan e^x)$
View full question & answer→Question 302 Marks
Differentiate the $\sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)$ w.r.t. x.
AnswerLet $y = \sin \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)$ $\therefore \frac{{dy}}{{dx}} = \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\frac{d}{{dx}}\left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\,\,\left[ {\because \frac{d}{{dx}}\sin f\left( x \right) = \cos f\left( x \right)\frac{d}{{dx}}f\left( x \right)} \right]$
$= \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\frac{1}{{1 + {{\left( {{e^{ - x}}} \right)}^2}}}\frac{d}{{dx}}{e^{ - x}}\,\,\left[ {\because \frac{d}{{dx}}{{\tan }^{ - 1}}f\left( x \right) = \frac{1}{{{{ 1+\left ({f\left( x \right)} \right)}^2}}}\frac{d}{{dx}}f\left( x \right)} \right]$
$= \cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)\frac{1}{{1 + {e^{ - 2x}}}}{e^{ - x}}\frac{d}{{dx}}\left( { - x} \right)$
$= - \frac{{{e^{ - x}}\cos \left( {{{\tan }^{ - 1}}{e^{ - x}}} \right)}}{{1 + {e^{ - 2x}}}}$
View full question & answer→Question 312 Marks
Differentiate $e^{x^{3}}$ w.r.t. x.
AnswerLet y = $e^{x^{3}}$
So, by using the chain rule, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(e^{x^{3}}\right)$
= $e^{x^{3}} \cdot \frac{d}{d x}\left(x^{3}\right)$
= $e^{x^{3}} \cdot 3 x^{2}$
= $3 x^{2} e^{x^{3}}$
View full question & answer→Question 322 Marks
Differentiate $e^{sin^{-1}x}$ w.r.t. x.
AnswerLet $y = e^{sin^{-1}x}$
Now, by using the chain rule, we get,
$\frac{d y}{d x}=\frac{d}{d x}\left(e^{\sin ^{-1} x}\right)$
$\Rightarrow \frac{d y}{d x}=e^{\sin ^{-1} x} \cdot \frac{d}{d x}\left(\sin ^{-1} x\right)$
= $e^{\sin ^{-1} x} \cdot \frac{1}{\sqrt{1-x^{2}}}$
= $\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}}$
Thus $\frac{d y}{d x}=\frac{e^{\sin ^{-1} x}}{\sqrt{1-x^{2}}}$
View full question & answer→Question 332 Marks
Differentiate the $\cos \left(\log x+e^x\right), x>0$ w.r.t. $x$.
AnswerLet y $= \cos(logx + e^x)$
$\therefore \frac{{dy}}{{dx}} = - \sin \left( {\log x + {e^x}} \right)\frac{d}{{dx}}\left( {\log x + {e^x}} \right)$
$ = - \sin \left( {\log x + {e^x}} \right).\left( {\frac{1}{x} + {e^x}} \right)$
View full question & answer→Question 342 Marks
Differentiate the $\frac{{{e^x}}}{{\sin x}}$ w.r.t x.
AnswerLet $y = \frac{{{e^x}}}{{\sin x}}$ $\therefore \frac{{dy}}{{dx}} = \frac{{\sin x\frac{d}{{dx}}{e^x} - {e^x}\frac{d}{{dx}}\sin x}}{{{{\sin }^2}x}}$ [By quotient rule]
$= \frac{{\sin x.{e^x} - {e^x}\cos x}}{{{{\sin }^2}x}}$
$= {e^x}\frac{{\left( {\sin x - \cos x} \right)}}{{{{\sin }^2}x}}$
View full question & answer→Question 352 Marks
Find $\frac{d y}{d x}$ if $y = \sin^{–1} \left(\frac{2 x}{1+x^{2}}\right)$
AnswerHere, y = $sin^{-1}(\frac{2x}{1+x^2})$
Let $x = tan A$
then, $A = \tan^{-1}x$
$\Rightarrow \frac{d A}{d x}=\frac{1}{1+x^{2}}$
y = $\sin ^{-1}\left(\frac{2 \tan A}{1+\tan ^{2} A}\right)$
Also, we know $\left[\sin 2 A=\frac{2 \tan A}{1+\tan ^{2} A}\right]$
$\Rightarrow y = sin^{-1} (sin2A)$
$\Rightarrow y = 2A$
$\Rightarrow$ $\frac{d y}{d x}=\frac{dy}{dA} \times\frac{dA}{dx}=2 \frac{d A}{d x}$ ...[By chain rule]
$\Rightarrow$ $\frac{d y}{d x}=\frac{2}{1+x^{2}}$
View full question & answer→Question 362 Marks
Find $\frac{dy}{dx}$ if $\sin^2 x + \cos^2 y = 1$
AnswerIt is given that $\sin^2 x + \cos^2 y = 1$
Differentiating both sides w.r.t. x, we get,
$\frac{d}{d x}\left(\sin ^{2} x+\cos ^{2} y\right)=\frac{d}{d x}(1)$
$\Rightarrow \frac{d}{d x}\left(\sin ^{2} x\right)+\frac{d}{d x}\left(\cos ^{2} y\right)=0$
$\Rightarrow 2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \cos y \cdot \frac{d}{d x}(\cos y)=0$
$\Rightarrow 2 \sin x \cos x+2 \cos y(-\sin y) \cdot \frac{d y}{d x}=0$
$\Rightarrow \sin 2 x-\sin 2 y \frac{d y}{d x}=0$
$\Rightarrow \frac{d y}{d x}=\frac{\sin 2 x}{\sin 2 y}$
View full question & answer→Question 372 Marks
Find $\frac{d y}{d x}$ if $\sin^2 y + \cos xy$ = $\kappa$
AnswerIt is given that $\sin^2 y + \cos xy$ = $\kappa$
Differentiating both sides w.r.t. $x$, we get,
$\frac{d}{d x}\left(\sin ^{2} y+\cos x y\right)=\frac{d}{d x}(\kappa)$
$\Rightarrow 2 \sin y \cos y \frac{d y}{d x}-\sin x y\left[y \frac{d}{d x}(x)+x \frac{d y}{d x}\right]=0$
$\Rightarrow 2 \sin y \cos y \frac{d y}{d x}-\sin x y\left[y \cdot 1+x \frac{d y}{d x}\right]=0$
$\Rightarrow 2 \sin y \cos y \frac{d y}{d x}-y \sin x y-x \sin x y \frac{d y}{d x}=0$
$\Rightarrow(2 \sin y \cos y-x \sin x y) \frac{d y}{d x}=y \sin x y$
$\Rightarrow(\sin 2 y-x \sin x y) \frac{d y}{d x}=y \sin x y$
$\Rightarrow \frac{d y}{d x}=\frac{\text { y sin xy }}{(\sin 2 y-x \sin x y)}$
View full question & answer→Question 382 Marks
Find $\frac{{dy}}{{dx}}$ if $x^3 + x^2y + xy^2 + y^3 = 81$.
Answerwe have,$x^3 + x^2y + xy^2 + y^3 = 81$
Differentiating both sides w.r.t to x,we get,
$3{x^2} + {x^2}.\frac{{dy}}{{dx}} + y.2x + x.2y\frac{{dy}}{{dx}} + {y^2}.1 + 3{y^2}\frac{{dy}}{{dx}} = 0$
$\left( {{x^2} + 2xy + 3{y^2}} \right)\frac{{dy}}{{dx}} = - 3{x^2} - 2xy - {y^2}$
$\frac{{dy}}{{dx}} = \frac{{ - \left( {3{x^2} + 2xy + {y^2}} \right)}}{{{x^2} + 2xy + 3{y^2}}}$
View full question & answer→Question 392 Marks
Find $\frac{{dy}}{{dx}}$ if ${x^2} + xy + {y^2} = 100$
AnswerGiven: ${x^2} + xy + {y^2} = 100$ $\Rightarrow \frac{d}{{dx}}{x^2} + \frac{d}{{dx}}xy + \frac{d}{{dx}}{y^2} = \frac{d}{{dx}}100$
$\Rightarrow 2x + \left( {x\frac{d}{{dx}}y + y\frac{d}{{dx}}x} \right) + 2y\frac{{dy}}{{dx}} = 0$
$ \Rightarrow 2x + x\frac{{dy}}{{dx}} + y + 2y\frac{{dy}}{{dx}} = 0$
$\Rightarrow \left( {x + 2y} \right)\frac{{dy}}{{dx}} = - 2x - y$
$ \Rightarrow \frac{{dy}}{{dx}} = \frac{{ - \left( {2x + y} \right)}}{{x + 2y}}$
View full question & answer→Question 402 Marks
Find $\frac{{dy}}{{dx}}$ if $xy + {y^2} = \tan x + y$
AnswerGiven: $xy + {y^2} = \tan x + y$ $\Rightarrow \frac{d}{{dx}}\left( {xy} \right) + \frac{d}{{dx}}\left( {{y^2}} \right) = \frac{d}{{dx}}\tan x + \frac{d}{{dx}}y$
$\Rightarrow x\frac{dy}{{dx}} + y.1 + 2y\frac{dy}{{dx}} = {\sec ^2}x + \frac{dy}{{dx}}$ [By Product Rule]
$\Rightarrow x\frac{dy}{{dx}} + 2y\frac{{dy}}{{dx}} - \frac{{dy}}{{dx}} = {\sec ^2}x - y$
$$$\Rightarrow \left( {x + 2y - 1} \right)\frac{{dy}}{{dx}} = {\sec ^2}x - y$
$$$\Rightarrow \frac{{dy}}{{dx}} = \frac{{{{\sec }^2}x - y}}{{x + 2y - 1}}$
View full question & answer→Question 412 Marks
Find $\frac{d y}{d x}$ if $ax + by^2 = \cos y$
AnswerIt is given that $ax + by^2 = \cos y$
Differentiating both sides w.r.t. x, we get,
$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}+\mathrm{by}^{2}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{cosy})$
$\Rightarrow \frac{d}{d x}(a x)+\frac{d}{d x}\left(b y^{2}\right)=\frac{d}{d x}(\cos y)$
$\Rightarrow a+b \frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\cos y)$
$\Rightarrow a+b \times 2 y \frac{d y}{d x}=-\sin y \frac{d y}{d x}$
$\Rightarrow(2 b y+\sin y) \frac{d y}{d x}=-a$
$\Rightarrow \frac{d y}{d x}=\frac{-a}{(2 b y+\sin y)}$
View full question & answer→Question 422 Marks
Find $\frac{{dy}}{{dx}}$ if $2x + 3y = \sin y$
AnswerGiven: $2x + 3y = \sin y$ $\Rightarrow \frac{d}{{dx}}\left( {2x} \right) + \frac{d}{{dx}}\left( {3y} \right) = \frac{d}{{dx}}\sin y$
$\Rightarrow 2 + 3\frac{{dy}}{{dx}} = \cos y\frac{{dy}}{{dx}}$
$\Rightarrow - \cos y\frac{{dy}}{{dx}} + 3\frac{{dy}}{{dx}} = - 2$
$\Rightarrow - \frac{{dy}}{{dx}}\left( {\cos y - 3} \right) = - 2$
$\Rightarrow \frac{{dy}}{{dx}} = \frac{2}{{\cos y - 3}}$
View full question & answer→Question 432 Marks
Find $\frac{dy}{dx}$ if $y=\sec ^{-1}\left(\frac{1}{2 x^{2}-1}\right), 0$
AnswerIt is given that $y=\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right)$
$\Rightarrow \sec y=\frac{1}{2 x^{2}+1}$
$\Rightarrow \cos y = 2x^2 + 1$
$\Rightarrow 2x^2 = 1 + \cos y$
$\Rightarrow 2x^2 = 2cos^2 \frac{y}{2}$
$\Rightarrow$ x = cos $\frac{y}{2}$
Differentiating w.r.t x, we get
$\frac{d}{d x}(x)=\frac{d}{d x}\left(\cos \frac{y}{2}\right)$
$\Rightarrow 1=-\sin \frac{\mathrm{y}}{2} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{y}}{2}\right)$
$\Rightarrow \frac{-1}{\sin \frac{y}{2}}=\frac{1}{2} \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos ^{2} \frac{y}{2}}}$
$\Rightarrow \frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}}$
View full question & answer→Question 442 Marks
Find $\frac{{dy}}{{dx}},$ if $y = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right),-\frac{{ 1}}{{\sqrt 2 }} < x < \frac{1}{{\sqrt 2 }}$
AnswerGiven: $y = {\sin ^{ - 1}}\left( {2x\sqrt {1 - {x^2}} } \right),-\frac{{ 1}}{{\sqrt 2 }} < x < \frac{1}{{\sqrt 2 }}$
Putting $x = \sin \theta$
$y = {\sin ^{ - 1}}\left( {2\sin \theta \sqrt {1 - {{\sin }^2}\theta } } \right)$
$= {\sin ^{ - 1}}\left( {2\sin \theta \sqrt {{{\cos }^2}\theta } } \right)$
$= {\sin ^{ - 1}}\left( {2\sin \theta \cos \theta } \right)$
$ = {\sin ^{ - 1}}\left( {\sin 2\theta } \right) = 2\theta = 2{\sin ^{ - 1}}x$
$\therefore \frac{{dy}}{{dx}} = 2.\frac{1}{{\sqrt {1 - {x^2}} }} = \frac{2}{{\sqrt {1 - {x^2}} }}$
View full question & answer→Question 452 Marks
Find $\frac{d y}{d x}$ if $y = \cos^{–1} \left(\frac{2 x}{1+x^{2}}\right),-1$
AnswerIt is given that y = $\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
$\Rightarrow \cos y=\frac{2 x}{1+x^{2}}$
Differentiating both sides w.r.t. x, we get,
$-\sin y \frac{d y}{d x}=\frac{\left(1+x^{2}\right) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow -\sqrt{1-\cos ^{2} y} \frac{d y}{d x}=\frac{\left(1+x^{2}\right) \times 2-2 x \cdot 2 x}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow \sqrt{1-\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)^{2}} \frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{-2\left(1-\mathrm{x}^{2}\right)}{~~~~~~\left(1+\mathrm{x}^{2}\right)^{2}}\right]$
$\Rightarrow \sqrt{\frac{\left(1+x^{2}\right)^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow \sqrt{\frac{\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow \frac{1-x^{2}}{1+x^{2}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}$
$\Rightarrow \frac{d y}{d x}=\frac{-2}{1+x^{2}}$
View full question & answer→Question 462 Marks
Find $\frac{{dy}}{{dx}},$ $y = {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right),0 < x < 1$
AnswerGiven: $y = {\sin ^{ - 1}}\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right),0 < x < 1$
Putting $x = \tan \theta $
$y = {\sin ^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)$
$= {\sin ^{ - 1}}\left( {\cos 2\theta } \right)$
$= {\sin ^{ - 1}}\sin \left( {\frac{\pi }{2} - 2\theta } \right) = \frac{\pi }{2} - 2\theta$
$= \frac{\pi }{2} - 2{\tan ^{ - 1}}x$
$\therefore \frac{{dy}}{{dx}} = 0 - 2.\frac{1}{{1 + {x^2}}} = \frac{{ - 2}}{{1 + {x^2}}}$
View full question & answer→Question 472 Marks
Find $\frac{d y}{d x}$, If $y = \cos^{–1} \left(\frac{1-x^{2}}{1+x^{2}}\right), 0 < x < 1$
AnswerIt is given that,
$y = \cos^{-1} \left(\frac{1-x^{2}}{1+x^{2}}\right)$
$\Rightarrow \cos y=\frac{1-x^{2}}{1+x^{2}}$
$\Rightarrow \frac{1-\tan ^{2} \frac{y}{2}}{1+\tan ^{2} \frac{y}{2}}=\frac{1-x^{2}}{1+x^{2}}$
On comparing both sides, we get
tan $\frac{y}{2}$ = x
Now, differentiating both sides, we get,
$\sec ^{2}\left(\frac{y}{2}\right) \cdot \frac{d}{d x}\left(\frac{y}{2}\right)=\frac{d}{d x}(x)$
$\Rightarrow \sec ^{2} \frac{y}{2} \times \frac{1}{2} \frac{d y}{d x}=1$
$\Rightarrow \frac{d y}{d x}=\frac{2}{\sec ^{2} \frac{y}{2}}$
$\Rightarrow \frac{d y}{d x}=\frac{2}{1+\tan ^{2} \frac{y}{2}}$
$\Rightarrow \frac{d y}{d x}=\frac{2}{1+x^{2}}$
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Find $\frac{{dy}}{{dx}},$ if $y = {\tan ^{ - 1}}\left( {\frac{{3x - {x^3}}}{{1 - 3{x^2}}}} \right), - \frac{1}{{\sqrt 3 }} < x < \frac{1}{{\sqrt 3 }}$
AnswerPut $x = \tan \theta $, where $\frac{{ - \pi }}{6} < \theta < \frac{\pi }{6}$ Therefore, $y = {\tan ^{ - 1}}\left( {\frac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }}} \right)$
$ = {\tan ^{ - 1}}\left( {\tan 3\theta } \right)$
$= 3\theta \left( {\because \frac{{ - \pi }}{2} < 3\theta < \frac{\pi }{2}} \right)$
$ = 3{\tan ^{ - 1}}x$
Hence, $\frac{{dy}}{{dx}} = \frac{3}{{1 + {x^2}}}$
View full question & answer→Question 492 Marks
Find $\frac{d y}{d x}$ if 2x + 3y = sin x
AnswerIt is given that 2x + 3y = sin x
Differentiating both sides w.r.t. x, we get,
$\frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\frac{d}{d x}(\sin x)$
$\Rightarrow 2+3 \frac{d y}{d x}$ = cos x
$\Rightarrow$ $3 \frac{d y}{d x}=\cos x-2$
$\Rightarrow \frac{d y}{d x}=\frac{\cos x-2}{3}$
View full question & answer→Question 502 Marks
Differentiate the function with respect to x : $\cos \left( {\sqrt x } \right)$
AnswerLet $y = \cos \left( {\sqrt x } \right)$ $\therefore \frac{{dy}}{{dx}} = - \sin \sqrt x. \frac{d}{{dx}}\sqrt x $
$ = - \sin \sqrt x .\frac{1}{2}{\left( x \right)^{\frac{{ - 1}}{2}}}$
$= \frac{{ - \sin \sqrt x }}{{2\sqrt x }}$
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