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$\cot ^{-1}\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)$
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Diffrentiate the following w. r. t. x$\tan ^{-1}\left(\frac{2 x^{\frac{5}{2}}}{1-x^5}\right)$
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Diffrentiate the following w. r. t. x$\sin ^{-1}\left(\frac{1-x^3}{1+x^3}\right)$
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Diffrentiate the following w. r. t. x$\sin ^{-1}\left(\frac{1-25 x^2}{1+25 x^2}\right)$
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Diffrentiate the following w. r. t. x$\sin ^{-1}\left(\frac{4^{x+\frac{1}{2}}}{1+2^{4 x}}\right)$
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Diffrentiate the following w. r. t. x$\operatorname{CoS}^{-1}\left(\frac{1-9^x}{1+9^x}\right)$
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Diffrentiate the following w. r. t. x$\cos ^{-1}\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)$
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