Question 13 Marks
If $y=A e^{m x}+B e^{n x}$, show that $y_2-(m+n) y_1+(m n) y=0$.
View full question & answer→Question 23 Marks
If $y=f(x)$ is a differentiable function of $x$, then show that $\frac{d^2 x}{d y^2}=-\left(\frac{d y}{d x}\right)^{-3} \cdot \frac{d^2 y}{d x^2}$
View full question & answer→Question 33 Marks
If $x =e^{\frac{x}{y}}$, then show that $\frac{d y}{d x}=\frac{x-y}{x \log x}$
View full question & answer→Question 43 Marks
If $\sin y=x \sin (a+y)$, then show that $\frac{d y}{d x}=\frac{\sin ^2(a+y)}{\sin a}$
View full question & answer→Question 53 Marks
Differentiate the following w.r.t. x:$\tan ^{-1}\left[\frac{\sqrt{x}(3-x)}{1-3 x}\right]$
View full question & answer→Question 63 Marks
Differentiate the following w.r.t. x:$\sin \left[2 \tan ^{-1}\left(\sqrt{\frac{1-x}{1+x}}\right)\right]$
View full question & answer→Question 73 Marks
Differentiate the following w. r. t. x.$y=\log \left[\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}-x}\right]$
Answer$
\begin{aligned}
y=\log \left[\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}-x}\right] & =\log \left[\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}-x} \times \frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}+x}\right] \\
& =\log \left[\frac{\left(\sqrt{x^2+a^2}+x\right)^2}{x^2+a^2-x^2}\right] \\
& =\log \left[\frac{\left(\sqrt{x^2+a^2}+x\right)^2}{a^2}\right] \\
& =\log \left(\sqrt{x^2+a^2}+x\right)^2-\log \left(a^2\right) \\
\therefore \quad & =2 \log \left(\sqrt{x^2+a^2}+x\right)-\log \left(a^2\right)
\end{aligned}
$
Differentiate $w, r, t . x$$\begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[2 \log \left(\sqrt{x^2+a^2}+x\right)-\log \left(a^2\right)\right] \\ & =2 \frac{d}{d x}\left[\log \left(\sqrt{x^2+a^2}+x\right)\right]-\frac{d}{d x}\left[\log \left(a^2\right)\right] \\ & =2 \times \frac{1}{\sqrt{x^2+a^2}+x} \cdot \frac{d}{d x}\left[\sqrt{x^2+a^2}+x\right]-0 \\ & =\frac{2}{\sqrt{x^2+a^2}+x} \cdot\left[\frac{1}{2 \sqrt{x^2+a^2}} \cdot \frac{d}{d x}\left(x^2+a^2\right)+1\right] \\ & =\frac{2}{\sqrt{x^2+a^2}+x} \cdot\left[\frac{1}{2 \sqrt{x^2+a^2}}(2 x)+1\right] \\ & =\frac{2}{\sqrt{x^2+a^2}+x} \cdot\left[\frac{x+\sqrt{x^2+a^2}}{\sqrt{x^2+a^2}}\right] \\ \frac{d y}{d x} & =\frac{2}{\sqrt{x^2+a^2}}\end{aligned}$
View full question & answer→Question 83 Marks
Differentiate the following w. r. t. x.$y=\log _3\left(\log _5 x\right)$
View full question & answer→Question 93 Marks
Find the $n^{\text {th }}$ derivative of the following : $\sin x$
AnswerLet $y=\sin x$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}(\sin x)=\cos x \\
& \frac{d y}{d x}=\sin \left(\frac{\pi}{2}+x\right)
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left[\sin \left(\frac{\pi}{2}+x\right)\right] \\
& \frac{d^2 y}{d x^2}=\cos \left(\frac{\pi}{2}+x\right) \frac{d}{d x}\left(\frac{\pi}{2}+x\right) \\
& \frac{d^2 y}{d x^2}=\sin \left(\frac{\pi}{2}+\frac{\pi}{2}+x\right)(1) \\
& \frac{d^2 y}{d x^2}=\sin \left(\frac{2 \pi}{2}+x\right)
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=\frac{d}{d x}\left[\sin \left(\frac{2 \pi}{2}+x\right)\right] \\
& \frac{d^3 y}{d x^3}=\cos \left(\frac{2 \pi}{2}+x\right) \frac{d}{d x}\left(\frac{2 \pi}{2}+x\right) \\
& \quad=\sin \left(\frac{\pi}{2}+\frac{2 \pi}{2}+x\right)(1) \\
& \frac{d^3 y}{d x^3}=\sin \left(\frac{3 \pi}{2}+x\right)
\end{aligned}
$
In general $n^{\text {th }}$ order derivative will be
$
\frac{d^n y}{d x^n}=\sin \left(\frac{n \pi}{2}+x\right)
$
View full question & answer→Question 103 Marks
Find the $n^{\text {th }}$ derivative of the following : $\log x$
AnswerLet $y=\log x$
Differentiate w.r.t. $x$
$
\frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x}
$
Differentiate $w . r . t . x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{1}{x}\right) \\
& \frac{d^2 y}{d x^2}=\frac{-1}{x^2}=\frac{(-1)^1}{x^2}
\end{aligned}
$
Differentiate $w . r . t . x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=(-1)^1 \frac{d}{d x}\left(\frac{1}{x^2}\right) \\
& \frac{d^3 y}{d x^3}=(-1)^1\left(\frac{-2}{x^3}\right)=\frac{(-1)^2 \cdot 1 \cdot 2}{x^3}
\end{aligned}
$
In general $n^{\text {th }}$ order derivative will be
$
\begin{aligned}
& \frac{d^n y}{d x^n}=\frac{(-1)^{n-1} \cdot 1 \cdot 2 \cdot 3 \ldots(n-1)}{x^n} \\
& \frac{d^n y}{d x^n}=\frac{(-1)^{n-1} \cdot(n-1) !}{x^n}
\end{aligned}
$
View full question & answer→Question 113 Marks
Find the $n^{\text {th }}$ derivative of the following : $\frac{1}{a x+b}$
AnswerLet $y=\frac{1}{a x+b}$
Differentiate w. r. t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(\frac{1}{a x+b}\right)=\frac{-1}{(a x+b)^2} \cdot \frac{d}{d x}(a x+b) \\
& \frac{d y}{d x}=\frac{(-1) \cdot a}{(a x+b)^2}
\end{aligned}
$
Differentiate w. r. t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=(-1)(a) \frac{d}{d x}\left(\frac{1}{(a x+b)^2}\right) \\
& \frac{d^2 y}{d x^2}=(-1)(a) \frac{-2}{(a x+b)^3} \cdot \frac{d}{d x}(a x+b) \\
& \frac{d^2 y}{d x^2}=\frac{(-1)^2 \cdot 2 \cdot 1 \cdot a^2}{(a x+b)^3}
\end{aligned}
$
Differentiate $w$. r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d^2 y}{d x^2}\right)=(-1)^2 \cdot 2 \cdot 1 \cdot a^2 \cdot \frac{d}{d x}\left(\frac{1}{(a x+b)^3}\right) \\
& \frac{d^3 y}{d x^3}=(-1)^2 \cdot 2 \cdot 1 \cdot a^2 \cdot \frac{-3}{(a x+b)^4} \cdot \frac{d}{d x}(a x+b) \\
& \frac{d^3 y}{d x^3}=\frac{(-1)^3 \cdot 3 \cdot 2 \cdot 1 \cdot a^3}{(a x+b)^4}
\end{aligned}
$
In general $n^{\text {th }}$ order derivative will be
$
\begin{aligned}
& \frac{d^n y}{d x^n}=\frac{(-1)^n \cdot n \cdot(n-1) \ldots 2 \cdot 1 \cdot a^n}{(a x+b)^{n+1}} \\
& \frac{d^n y}{d x^n}=\frac{(-1)^n \cdot n ! \cdot a^n}{(a x+b)^{n+1}}
\end{aligned}
$
View full question & answer→Question 123 Marks
If $x=\sin t, y=e^{m t}$ then show that $\left(1-x^2\right) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0$.
AnswerGiven that $x=\sin t \quad \therefore \quad t=\sin ^{-1} x$
and $y=e^{m t} \quad \therefore \quad y=e^{m \sin ^{-1} x}$
Differentiate w.r.t. $x$
$\frac{d y}{d x}=\frac{d}{d x}\left(e^{m \sin ^{-1} x}\right)=e^{m \sin ^{-1} x} \cdot m \frac{d}{d x}\left(\sin ^{-1} x\right)$
$\frac{d y}{d x}=\frac{m \cdot e^{m \sin ^{-1} x}}{\sqrt{1-x^2}}$
$\sqrt{1-x^2} \frac{d y}{d x}=m y$
$\ldots[$ From (I)]
Squaring both sides
$
\left(1-x^2\right) \cdot\left(\frac{d y}{d x}\right)^2=m^2 y^2
$
Differentiate $w \cdot r, t . x$
$
\begin{aligned}
& \left(1-x^2\right) \frac{d}{d x}\left(\frac{d y}{d x}\right)^2+\left(\frac{d y}{d x}\right)^2 \frac{d}{d x}\left(1-x^2\right)=m^2 \frac{d}{d x}\left(y^2\right) \\
& \left(1-x^2\right) \cdot 2\left(\frac{d y}{d x}\right) \cdot \frac{d}{d x} \cdot\left(\frac{d y}{d x}\right)+\left(\frac{d y}{d x}\right)^2(-2 x)=m^2(2 y) \frac{d y}{d x}
\end{aligned}
$
$
2\left(1-x^2\right) \cdot \frac{d y}{d x} \cdot \frac{d^2 y}{d x^2}-2 x\left(\frac{d y}{d x}\right)^2=2 m^2 y \frac{d y}{d x}
$
Dividing throughout by $2 \frac{d y}{d x}$ we get,
$
\begin{gathered}
\left(1-x^2\right) \cdot \frac{d^2 y}{d x^2}-x \frac{d y}{d x}=m^2 y \\
\therefore \quad\left(1-x^2\right) \cdot \frac{d^2 y}{d x^2}-x \frac{d y}{d x}-m^2 y=0
\end{gathered}
$
View full question & answer→Question 133 Marks
Find the second order derivative of the following : $e^{2 x} \sin 3 x$
AnswerLet $y=e^{2 x} \sin 3 x$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(e^{2 x} \sin 3 x\right)=e^{2 x} \frac{d}{d x}(\sin 3 x)+\sin 3 x \frac{d}{d x}\left(e^{2 x}\right) \\
& \frac{d y}{d x}=e^{2 x}(\cos 3 x)(3)+\sin 3 x\left(e^{2 x}\right)(2) \\
& \frac{d y}{d x}=e^{2 x}(3 \cos 3 x+2 \sin 3 x)
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left[e^{2 x}(3 \cos 3 x+2 \sin 3 x)\right] \\
& \frac{d^2 y}{d x^2}=e^{2 x} \frac{d}{d x}(3 \cos 3 x+2 \sin 3 x)+(3 \cos 3 x+2 \sin 3 x) \frac{d}{d x}\left(e^{2 x}\right) \\
& =e^{2 x}[3(-\sin 3 x)(3)+2(\cos 3 x)(3)]+(3 \cos 3 x+2 \sin 3 x) e^{2 x}(2) \\
& =e^{2 x}[-9 \sin 3 x+6 \cos 3 x+6 \cos 3 x+4 \sin 3 x] \\
& \frac{d^2 y}{d x^2}=e^{2 x}[12 \cos 3 x-5 \sin 3 x] \\
&
\end{aligned}
$
View full question & answer→Question 143 Marks
Find the second order derivative of the following : $x^2 e^x$
AnswerLet $y=x^2 e^x$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{d}{d x}\left(x^2 e^x\right) \\
& \frac{d y}{d x}=x^2 \frac{d}{d x}\left(e^x\right)+e^x \frac{d}{d x}\left(x^2\right) \\
& \frac{d y}{d x}=x^2 e^x+2 x e^x=e^x\left(x^2+2 x\right)
\end{aligned}
$
Differentiate w.r. $t . x$
$
\begin{aligned}
& \frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left[e^x\left(x^2+2 x\right)\right] \\
& \frac{d^2 y}{d x^2}=e^x \frac{d}{d x}\left(x^2+2 x\right)+\left(x^2+2 x\right) \frac{d}{d x}\left(e^x\right) \\
& =e^x(2 x+2)+\left(x^2+2 x\right)\left(e^r\right) \\
& =\left(x^2+4 x+2\right) e^x \\
& \frac{d^2 y}{d x^2}=\left(x^2+4 x+2\right) e^x \\
\end{aligned}
$
View full question & answer→Question 153 Marks
Find the derivative of $\cos ^{-1} x$ w. r. t. $\sqrt{1-x^2}$.
AnswerLet $u=\cos ^{-1} x$ and $v=\sqrt{1-x^2}$, then we have to find $\frac{d u}{d v}$.
i.e. $\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}}$
Now, $u=\cos ^{-1} x$
Differentiate w.r. t. $x$
$\frac{d u}{d x}=\frac{d}{d x}\left(\cos ^{-1} x\right)=-\frac{1}{\sqrt{1-x^2}} \quad \ldots$ (II)
And, $v=\sqrt{1-x^2}$
Differentiate $w . r . t . x$
$
\begin{aligned}
\frac{d v}{d x} & =\frac{d}{d x}\left(\sqrt{1-x^2}\right)=\frac{1}{2 \sqrt{1-x^2}} \cdot \frac{d}{d x}\left(1-x^2\right)=\frac{1}{2 \sqrt{1-x^2}} \cdot(-2 x) \\
\frac{d v}{d x}=-\frac{x}{\sqrt{1-x^2}} & \ldots \text { (III) }
\end{aligned}
$
Substituting (II) and (III) in (I) we get,
$
\frac{d u}{d v}=\frac{-\frac{1}{\sqrt{1-x^2}}}{-\frac{x}{\sqrt{1-x^2}}} \quad \therefore \quad \frac{d u}{d v}=\frac{1}{x}
$
View full question & answer→Question 163 Marks
If $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$, then show that $\frac{d y}{d x}=\frac{b^2 x}{a^2 y}$.
AnswerGiven that, $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$ i.e. $\frac{x}{a}=t-\frac{1}{t} \ldots$
(I) and $\frac{y}{b}=t+\frac{1}{t} \ldots$
Square of (I) - Square of (II) gives,
$
\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{b^2}=\left(t-\frac{1}{t}\right)^2-\left(t+\frac{1}{t}\right)^2 \\
& =t^2-2+\frac{1}{t^2}-t^2-2-\frac{1}{t^2} \\
& \therefore \quad \frac{x^2}{a^2}-\frac{y^2}{a^2}=-4 \\
& \text { Differentiate w.r.t. } x \\
& \frac{1}{a^2} \cdot \frac{d}{d x}\left(x^2\right)-\frac{1}{b^2} \cdot \frac{d}{d x}\left(y^2\right)=\frac{d}{d x}(-4) \\
& \frac{1}{a^2}(2 x)-\frac{1}{b^2}(2 y) \cdot \frac{d}{d x}=0 \\
& \frac{1}{a^2}(2 x)-\frac{1}{b^2}(2 y) \cdot \frac{d y}{d x}=0 \\
& \frac{2 y}{b^2} \cdot \frac{d y}{d x}=\frac{2 x}{a^2} \Rightarrow \frac{d y}{d x}=\frac{b^2 x}{a^2 y} \\
& \therefore \quad \frac{d y}{d x}=\frac{b^2 x}{a^2 y}
\end{aligned}
$
View full question & answer→Question 173 Marks
If $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$, then show that $x^3 y \frac{d y}{d x}=1$.
AnswerGiven that, $x^4+y^4=t^2+\frac{1}{t^2}$
And
$
x^2+y^2=t+\frac{1}{t}
$
Squaring both sides,
$
\begin{aligned}
& \left(x^2+y^2\right)^2=\left(t+\frac{1}{t}\right)^2 \\
& x^4+2 x^2 y^2+y^4=t^2+2+\frac{1}{t^2} \\
& x^4+2 x^2 y^2+y^4=x^4+y^4+2 \\
& 2 x^2 y^2=2 \quad \therefore \quad x^2 y^2=1
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(x^2 y^2\right)=\frac{d}{d x}(1) \\
& x^2 \frac{d}{d x}\left(y^2\right)+y^2 \frac{d}{d x}\left(x^2\right)=0 \\
& x^2(2 y) \frac{d y}{d x}+y^2(2 x)=0 \\
& 2 x^2 y \frac{d y}{d x}=-2 x y^2 \Rightarrow \frac{d y}{d x}=-\frac{2 x y^2}{2 x^2 y} \\
& \qquad \frac{d y}{d x}=-\frac{x\left(-\frac{1}{x^2}\right)}{x^2 y} \ldots[\text { From (II)] } \\
& \quad \frac{d y}{d x}=\frac{1}{x^3 y} \quad \therefore \quad x^3 y \frac{d y}{d x}=1
\end{aligned}
$
View full question & answer→Question 183 Marks
Find $\frac{d y}{d x}$ if : $x=3 \cos t-2 \cos ^3 t, y=3 \sin t-2 \sin ^3 t$, at $t=\frac{\pi}{6}$
AnswerGiven, $y=3 \sin t-2 \sin ^3 t$
Differentiate w.r.t.t
$
\begin{aligned}
\frac{d y}{d t} & =\frac{d}{d t}\left(3 \sin t-2 \sin ^3 t\right) \\
& =3 \frac{d}{d t}(\sin t)-2(\sin t)^3 \\
& =3 \cos t-2(3) \sin ^2 t \frac{d}{d t}(\sin t) \\
& =3 \cos t-6 \sin ^2 t(\cos t) \\
& =3 \cos t\left(1-2 \sin ^2 t\right) \\
\frac{d y}{d t} & =3 \cos t \cos 2 t
\end{aligned}
$
And, $x=3 \cos t-2 \cos ^3 t$
Differentiate w.r.t.t
$
\begin{aligned}
\frac{d x}{d t} & =\frac{d}{d t}\left(3 \cos t-2 \cos ^3 t\right) \\
& =3 \frac{d}{d t}(\cos t)-2 \frac{d}{d t}\left(\cos ^3 t\right) \\
& =3(-\sin t)-2(3) \cos ^2 t \frac{d}{d t}(\cos t) \\
& =-3 \sin t-6 \cos ^2 t(-\sin t) \\
& =6 \cos ^2 t \sin t-3 \sin t \\
& =3 \sin t\left(2 \cos ^2 t-1\right) \\
\frac{d x}{d t} & =3 \sin t \cos 2 t
\end{aligned}
$
$
\therefore \quad \frac{d y}{d x}=-\cot t
$
At $t=\frac{\pi}{6}$, we get
$
\left(\frac{d y}{d x}\right)_{t=\frac{\pi}{6}}=-\cot \left(\frac{\pi}{6}\right)=\sqrt{3}
$
View full question & answer→Question 193 Marks
Find $\frac{d y}{d x}$ if : $x=t+\frac{1}{t}, y=\frac{1}{t^2}$, at $t=\frac{1}{2}$
AnswerGiven, $y=\frac{1}{t^2}$
Differentiate $w . r . t . t$
$\frac{d y}{d t}=\frac{d}{d t}\left(\frac{1}{t^2}\right)$
$\frac{d y}{d t}=-\frac{2}{t^3}$
And, $x=t+\frac{1}{t}$
Differentiate $w . r . t . t$
$
\begin{aligned}
& \frac{d x}{d t}=\frac{d}{d t}\left(t+\frac{1}{t}\right)=1-\frac{1}{t^2} \\
& \frac{d x}{d t}=-\frac{t^2-1}{t^2}
\end{aligned}
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-\frac{2}{t^3}}{\frac{t^2-1}{t^2}} \quad \ldots$ [From (I) and (II)]
$
\therefore \quad \frac{d y}{d x}=-\frac{2}{t\left(t^2-1\right)}
$
At $t=\frac{1}{2}$, we get
$
\begin{aligned}
\left(\frac{d y}{d x}\right)_{t=\frac{1}{2}} & =-\frac{2}{\left(\frac{1}{2}\right)\left[\left(\frac{1}{2}\right)^2-1\right]} \\
& =-\frac{2}{\left(\frac{1}{2}\right)\left(\frac{1}{4}-1\right)} \\
\left(\frac{d y}{d x}\right)_{t=\frac{1}{2}} & =-\frac{2}{\left(\frac{1}{2}\right)\left(-\frac{3}{4}\right)} \\
\left(\frac{d y}{d x}\right)_{t=\frac{1}{2}} & =\frac{16}{3}
\end{aligned}
$
View full question & answer→Question 203 Marks
Find $\frac{d y}{d x}$ if : $x=\sec ^2 \theta, y=\tan ^3 \theta$, at $\theta=\frac{\pi}{3}$
AnswerGiven, $y=\tan ^3 \theta$
Differentiate w.r. t. $\theta$
$\frac{d y}{d \theta}=\frac{d}{d \theta}(\tan \theta)^3=3 \tan ^2 \theta \frac{d}{d \theta}(\tan \theta) \quad \therefore \quad \frac{d y}{d \theta}=3 \tan ^2 \theta \cdot \sec ^2 \theta$
And, $x=\sec ^2 \theta$
Differentiate w.r.t. $\theta$
$
\begin{aligned}
& \frac{d x}{d \theta}=\frac{d}{d \theta}\left(\sec ^2 \theta\right)=2 \sec \theta \cdot \frac{d}{d \theta}(\sec \theta) \\
& \frac{d x}{d \theta}=2 \sec \theta \cdot \sec \theta \tan \theta=2 \sec ^2 \theta \cdot \tan \theta
\end{aligned}
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d \theta}}{\frac{d x}{d \theta}}=\frac{3 \tan ^2 \theta \cdot \sec ^2 \theta}{2 \sec ^2 \theta \cdot \tan \theta}$
$\ldots[$ From (I) and (II)]
$
\therefore \quad \frac{d y}{d x}=\frac{3}{2} \tan \theta
$
At $\theta=\frac{\pi}{3}$, we get
$
\left(\frac{d y}{d x}\right)_{\theta=\frac{\pi}{3}}=\frac{3}{2} \tan \left(\frac{\pi}{3}\right)=\frac{3 \sqrt{3}}{2}
$
View full question & answer→Question 213 Marks
Find $\frac{d y}{d x}$ if : $x=\sqrt{1-t^2}, y=\sin ^{-1} t$
AnswerGiven, $y=\sin ^{-1} t$
Differentiate w.r.t.t
$
\begin{aligned}
& \frac{d y}{d t}=\frac{d}{d t}\left(\sin ^{-1} t\right)=\frac{1}{\sqrt{1-t^2}} \\
& \frac{d y}{d t}=\frac{1}{\sqrt{1-t^2}} \\
& \text { And, } x=\sqrt{1-t^2}
\end{aligned}
$
And, $x=\sqrt{1-t^2}$
Differentiate w.r.t.t
$
\begin{aligned}
& \frac{d x}{d t}=\frac{d}{d t}\left(\sqrt{1-t^2}\right)=\frac{1}{2 \sqrt{1-t^2}} \cdot \frac{d}{d t}\left(1-t^2\right) \\
& \left.\frac{d x}{d t}=\frac{1}{2 \sqrt{1-t^2}} \cdot(-2 t)=-\frac{t}{\sqrt{1-t^2}} \ldots \ldots(1)\right)
\end{aligned}
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{\frac{1}{\sqrt{1-t^2}}}{-\frac{t}{\sqrt{1-t^2}}} \ldots[$ From (I) and (II) $]$
$
\therefore \quad \frac{d y}{d x}=-\frac{1}{t}
$
View full question & answer→Question 223 Marks
Find $\frac{d y}{d x}$ if : $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$
AnswerGiven, $y=a(1-\cos \theta)$
Differentiate w.r. t. $\theta$
$
\begin{aligned}
& \frac{d y}{d \theta}=a \frac{d}{d \theta}[(1-\cos \theta)]=a[0-(-\sin \theta)] \\
& \frac{d y}{d t}=a \sin \theta
\end{aligned}
$
And, $x=a(\theta+\sin \theta)$
Differentiate w.r.t. $\theta$
$
\begin{aligned}
& \frac{d x}{d t}=a \frac{d}{d \theta}(\theta+\sin \theta)=a(1+\cos \theta) \\
& \frac{d x}{d t}=a(1+\cos \theta)
\end{aligned}
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{a \sin \theta}{a(1+\cos \theta)} \ldots[$ From (I) and
$
\therefore \quad \frac{d y}{d x}=\frac{2 \sin \left(\frac{\theta}{2}\right) \cdot \cos \left(\frac{\theta}{2}\right)}{2 \cos ^2\left(\frac{\theta}{2}\right)}=\tan \left(\frac{\theta}{2}\right)
$
View full question & answer→Question 233 Marks
If $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}}$, then show that $\frac{d y}{d x}=\frac{\sec ^2 x}{2 y-1}$.
AnswerGiven that : $y=\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}}$
Squaring both sides, we get
$
\begin{aligned}
& y^2=\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}, \text { which is same as } \\
& y^2=\tan x+\sqrt{\tan x+\sqrt{\tan x+\sqrt{\tan x+\ldots \infty}}} \\
& y^2=\tan x+y \quad \ldots \ldots[\text { From (I) ] }
\end{aligned}
$
[From (I) ]
Differentiate $w . r . t . x$
$
\frac{d}{d x}\left(y^2\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x}
$
$
\begin{aligned}
& 2 y \frac{d y}{d x}-\frac{d y}{d x}=\sec ^2 x \\
& (2 y-1) \frac{d y}{d x}=\sec ^2 x \\
\therefore \quad & \frac{d y}{d x}=\frac{\sec ^2 x}{2 y-1}
\end{aligned}
$
View full question & answer→Question 243 Marks
If $\sec ^{-1}\left(\frac{x^3+y^3}{x^3-y^3}\right)=2 a$, then show that $\frac{d y}{d x}=\frac{x^2 \tan ^2 a}{y^2}$, where $a$ is a constant.
AnswerGiven that : $\sec ^{-1}\left(\frac{x^3+y^3}{x^3-y^3}\right)=2 a \quad \ldots$. [We will not eliminate $a$, as answer contains $a$ ]
$
\begin{aligned}
\therefore \quad & \cos ^{-1}\left(\frac{x^3-y^3}{x^3+y^3}\right)=2 a \\
& \frac{x^3-y^3}{x^3+y^3}=\cos 2 a \\
& x^3-y^3=x^3 \cos 2 a+y^3 \cos 2 a \\
& x^3-x^3 \cos 2 a=y^3 \cos 2 a+y^3 \\
& x^3(1-\cos 2 a)=y^3(1+\cos 2 a) \\
& y^3=\left(\frac{1-\cos 2 a}{1+\cos 2 a}\right) x^3 \\
& y^3=\left(\frac{2 \sin ^2 a}{2 \cos ^2 a}\right) x^3 \\
& y^3=\left(\tan ^2 a\right) x^3
\end{aligned}
$
Differentiate w.r.t. $x$
$
\begin{aligned}
& \frac{d}{d x}\left(y^3\right)=\left(\tan ^2 a\right) \frac{d}{d x}\left(x^3\right) \\
& 3 y^2 \frac{d y}{d x}=\left(\tan ^2 a\right) 3 x^2 \\
\therefore \quad & \frac{d y}{d x}=\frac{x^2 \tan ^2 a}{y^2}
\end{aligned}
$
View full question & answer→Question 253 Marks
If $\sin \left(\frac{p x^m-q y^m}{p x^m+q y^m}\right)=r$, then show that $\frac{d y}{d x}=\frac{y}{x}$, where $r$ is a constant.
AnswerGiven that : $\sin \left(\frac{p x^m-q y^m}{p x^m+q y^m}\right)=r$
$
\begin{aligned}
& \frac{p x^m-q y^m}{p x^m+q y^m}=\sin ^{-1} r \\
& \frac{p x^m-q y^m}{p x^m+q y^m}=t \quad \ldots \ldots\left[\text { Let } t=\sin ^{-1} r\right] \\
& p x^m-q y^m=p t x^m+q t y^m \\
& p x^m-p t x^m=q y^m+q t y^m \\
& p(1-t) x^m=q(1+t) y^m \\
& y^m=\left(\frac{p(1-t)}{q(1+t)}\right) x^m \\
& y^m=s \cdot x^m \\
& \text { Differentiate w.r.t. } x \\
& \frac{d}{d x}\left(y^m\right)=s \frac{d}{d x}\left(x^m\right) \\
& m y^{m-1} \frac{d y}{d x}=s \cdot m x^{m-1} \\
& \frac{d y}{d x}=s \cdot \frac{x^{m-1}}{y^{m-1}} x^{m-1} \\
& \frac{d y}{d x}=\frac{y^m}{x^m} \times \frac{x^{m-1}}{y^{m-1}} \quad \ldots \ldots[\text { From (I) }] \\
& \therefore \quad \frac{d y}{d x}=\frac{y}{x} \\
\end{aligned}
$
View full question & answer→Question 263 Marks
Find $\frac{d y}{d x}$ if : $x^2+e^{x y}=y^2+\log (x+y)$
AnswerGiven that : $x^2+e^{x y}=y^2+\log (x+y)$
Recall that $\frac{d}{d x} g(f(x))=g^{\prime}(f(x)) \cdot \frac{d}{d x} f(x)$
Differentiate $w . r . t . x$.
$
\begin{aligned}
& \frac{d}{d x}\left(x^2\right)+\frac{d}{d x}\left[e^{x y}\right]=\frac{d}{d x}\left(y^2\right)+\frac{d}{d x}[\log (x+y)] \\
& 2 x+e^{x y} \frac{d}{d x}(x y)=2 y \frac{d y}{d x}+\frac{1}{x+y} \frac{d}{d x}(x+y) \\
& 2 x+e^{x y}\left[x \frac{d y}{d x}+y(1)\right]=2 y \frac{d y}{d x}+\frac{1}{x+y}\left[1+\frac{d y}{d x}\right] \\
& 2 x+x e^{x y} \frac{d y}{d x}+y e^{x y}=2 y \frac{d y}{d x}+\frac{1}{x+y}+\frac{1}{x+y} \cdot \frac{d y}{d x} \\
& 2 x+y e^{x y}-\frac{1}{x+y}=2 y \frac{d y}{d x}-x e^{x y} \frac{d y}{d x}+\frac{1}{x+y} \cdot \frac{d y}{d x} \\
& 2 x+y e^{x y}-\frac{1}{x+y}=\left[2 y-x e^{x y}+\frac{1}{x+y}\right] \frac{d y}{d x} \\
& \frac{2 x(x+y)+y e^{x y}(x+y)-1}{x+y}=\left[\frac{2 y(x+y)-x e^{x y}(x+y)+1}{x+y}\right] \frac{d y}{d x} \\
\therefore & \frac{d y}{d x}=\frac{2 x(x+y)+y e^{y y}(x+y)-1}{2 y(x+y)-x e^{y y}(x+y)+1}
\end{aligned}
$
View full question & answer→Question 273 Marks
Find $\frac{d y}{d x}$ if : $y^3+\cos (x y)=x^2-\sin (x+y)$
AnswerGiven that: $y^3+\cos (x y)=x^2-\sin (x+y)$
Differentiate $w . r . t . x$.
$
\begin{aligned}
& \frac{d}{d x}\left(y^3\right)+\frac{d}{d x}[\cos (x y)]=\frac{d}{d x}\left(x^2\right)-\frac{d}{d x}[\sin (x+y)] \\
& 3 y^2 \frac{d}{d x}(y)-\sin (x y) \frac{d}{d x}(x y)=2 x-\cos (x+y) \frac{d}{d x}(x+y) \\
& 3 y^2 \frac{d y}{d x}-\sin (x y)\left[x \frac{d y}{d x}+y(1)\right]=2 x-\cos (x+y)\left[1+\frac{d y}{d x}\right] \\
& 3 y^2 \frac{d y}{d x}-x \sin (x y) \frac{d y}{d x}-y \sin (x y)=2 x-\cos (x+y)-\cos (x+y) \frac{d y}{d x} \\
& 3 y^2 \frac{d y}{d x}-x \sin (x y) \frac{d y}{d x}+\cos (x+y) \frac{d y}{d x}=2 x+y \sin (x y)-\cos (x+y) \\
& {\left[3 y^2-x \sin (x y)+\cos (x+y)\right] \frac{d y}{d x}=2 x+y \sin (x y)-\cos (x+y) } \\
\therefore & \frac{d y}{d x}=\frac{2 x+y \sin (x y)-\cos (x+y)}{3 y^2-x \sin (x y)+\cos (x+y)}
\end{aligned}
$
View full question & answer→Question 283 Marks
Find $\frac{d y}{d x}$ if : $x^5+x y^3+x^2 y+y^4=4$
AnswerGiven that: $x^5+x y^3+x^2 y+y^4=4$
Differentiate $w, r, t, x$.
$
\begin{aligned}
& \frac{d}{d x}\left(x^5\right)+\frac{d}{d x}\left(x y^3\right)+\frac{d}{d x}\left(x^2 y\right)+\frac{d}{d x}\left(y^4\right)=\frac{d}{d x}(4) \\
& 5 x^4+x \frac{d}{d x}\left(y^3\right)+y^3 \frac{d}{d x}(x)+x^2 \frac{d}{d x}(y)+y \frac{d}{d x}\left(x^2\right)+4 y^3 \frac{d}{d x}(y)=0 \\
& 5 x^4+x\left(3 y^2\right) \frac{d y}{d x}+y^3(1)+x^2 \frac{d y}{d x}+y(2 x)+4 y^3 \frac{d y}{d x}=0 \\
& x^2 \frac{d y}{d x}+3 x y^2 \frac{d y}{d x}+4 y^3 \frac{d y}{d x}=-5 x^4-2 x y-y^3 \\
& \left(x^2+3 x y^2+4 y^3\right) \frac{d y}{d x}=-\left(5 x^4+2 x y+y^3\right) \\
\therefore \quad & \frac{d y}{d x}=-\frac{5 x^4+2 x y+y^3}{x^2+3 x y^2+4 y^3}
\end{aligned}
$
View full question & answer→Question 293 Marks
Differentiate the following w. r. t. x.$\left(\frac{\left(x^2+3\right)^2 \sqrt[3]{\left(x^3+5\right)^2}}{\sqrt{\left(2 x^2+1\right)^3}}\right)$
AnswerLet $\begin{aligned} y & =\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1+2-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+x-2\right)}\right) \\ & =\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+4 x-3 x-2\right)}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-[2 x(3 x+2)-(3 x+2)]}\right) \\ & =\tan ^{-1}\left(\frac{5 x+1}{1-(3 x+2)(2 x-1)}\right)=\tan ^{-1}\left(\frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)}\right)\end{aligned}$$
y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)
$
Differentiate $w . r . t . x$.
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)\right] \\
& =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)\right]+\frac{d}{d x}\left[\tan ^{-1}(2 x-1)\right] \\
& =\frac{1}{1+(3 x+2)^2} \cdot \frac{d}{d x}(3 x+2)+\frac{1}{1+(2 x-1)^2} \cdot \frac{d}{d x}(2 x-1) \\
\therefore \quad \frac{d y}{d x} & =\frac{3}{1+(3 x+2)^2}+\frac{2}{1+(2 x-1)^2}
\end{aligned}
$
View full question & answer→Question 303 Marks
Differentiate the following w. r. t. x.$\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$
Answer$
\text { (iv) } \begin{aligned}
\text { Let } y & =\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1+2-x-6 x^2}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+x-2\right)}\right) \\
& =\tan ^{-1}\left(\frac{5 x+1}{1-\left(6 x^2+4 x-3 x-2\right)}\right)=\tan ^{-1}\left(\frac{5 x+1}{1-[2 x(3 x+2)-(3 x+2)]}\right) \\
& =\tan ^{-1}\left(\frac{5 x+1}{1-(3 x+2)(2 x-1)}\right)=\tan ^{-1}\left(\frac{(3 x+2)+(2 x-1)}{1-(3 x+2)(2 x-1)}\right)
\end{aligned}
$
$
\begin{aligned}
& y=\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1) \\
& \text { Differentiate w.r.t. } x \text {. } \\
& \frac{d y}{d x}=\frac{d}{d x}\left[\tan ^{-1}(3 x+2)+\tan ^{-1}(2 x-1)\right] \\
& =\frac{d}{d x}\left[\tan ^{-1}(3 x+2)\right]+\frac{d}{d x}\left[\tan ^{-1}(2 x-1)\right] \\
& =\frac{1}{1+(3 x+2)^2} \cdot \frac{d}{d x}(3 x+2)+\frac{1}{1+(2 x-1)^2} \cdot \frac{d}{d x}(2 x-1) \\
& \therefore \quad \frac{d y}{d x}=\frac{3}{1+(3 x+2)^2}+\frac{2}{1+(2 x-1)^2} \\
\end{aligned}
$
View full question & answer→Question 313 Marks
Differentiate the following w. r. t. x.$\cot ^{-1}\left(\frac{b \sin x-a \cos x}{a \sin x+b \cos x}\right)$
View full question & answer→Question 323 Marks
Differentiate the following w. r. t. x.$\tan ^{-1}\left(\frac{7 x}{1-12 x^2}\right)$
View full question & answer→Question 333 Marks
Differentiate the following w. r. t. x.$\sin ^{-1}\left(\frac{5 \sqrt{1-x^2}-12 x}{13}\right)$
View full question & answer→Question 343 Marks
Differentiate the following w. r. t. x.$\tan ^{-1}\left(\sqrt{\frac{3-x}{3+x}}\right)$
View full question & answer→Question 353 Marks
Differentiate the following w. r. t. x.$\cot ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$
AnswerLet $y=\cot ^{-1}\left(\frac{\cos x}{1+\sin x}\right)=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right)$
$=\tan ^{-1}\left(\frac{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]^2}{\cos ^2\left(\frac{x}{2}\right)-\sin ^2\left(\frac{x}{2}\right)}\right)=\tan ^{-1}\left(\frac{\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]^2}{\left[\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)\right]\left[\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)\right]}\right)$
$=\tan ^{-1}\left(\frac{\cos \left(\frac{x}{2}\right)+\sin \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right)-\sin \left(\frac{x}{2}\right)}\right)=\tan ^{-1}\left(\frac{1+\tan \left(\frac{x}{2}\right)}{1-\tan \left(\frac{x}{2}\right)}\right) \ldots$. Divide Numerator \& Denominator by $\cos \left(\frac{x}{2}\right)$
$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right]$
$\therefore y=\frac{\pi}{4}+\frac{x}{2}$
Differentiate $w . r . t . x$.
$
\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{4}+\frac{x}{2}\right)=0+\frac{1}{2} \quad \therefore \frac{d y}{d x}=\frac{1}{2}
$
View full question & answer→Question 363 Marks
Find the nth derivative of the following:
log(ax + b)
Question 373 Marks
If $y = x +\tan x$, show that $\cos ^2 x \cdot \frac{d^2 y}{d x^2}-2 y+2 x=0$
View full question & answer→Question 383 Marks
Differentiate $\tan ^{-1}\left(\frac{\cos x}{1+\sin x}\right)$ w.r.t. $\sec ^{-1} x$.
View full question & answer→Question 393 Marks
Differentiate 3x w.r.t. $\log_{x_ 3. }$
View full question & answer→Question 403 Marks
Differentiate x sin x w.r.t tan x.
Question 413 Marks
If $x =\log \left(1+ t ^2\right), y = t -\tan ^{-1} t$, show that $\frac{d y}{d x}=\frac{\sqrt{e^x-1}}{2}$
View full question & answer→Question 423 Marks
If $x=2 \cos ^4(t+3), y=3 \sin ^4(t+3)$, show that $\frac{d y}{d x}=-\sqrt{\frac{3 y}{2 x}}$
View full question & answer→Question 433 Marks
If $x = a \cos ^3 t , y = a \sin ^3 t$, then show that $\frac{d y}{d x}=-\left(\frac{y}{x}\right)^{\frac{1}{3}}$
View full question & answer→Question 443 Marks
If $x =\frac{t+1}{t-1}, y =\frac{1-t}{t+1}$, then show that $y 2-\frac{d y}{d x}=0$.
View full question & answer→Question 453 Marks
If $x =e^{ sin 3 t}, y =e^{\cos 3 t}$, then show that $\frac{d y}{d x}=-\frac{y \log x}{x \log y}$
View full question & answer→Question 463 Marks
Find $\frac{d y}{d x}$ if$x=\cos ^{-1}\left(4 t^3-3 t\right), y=\tan ^{-1}\left(\frac{\sqrt{1-t^2}}{t}\right)$
View full question & answer→Question 473 Marks
Find $\frac{d y}{d x}$ if$x=\cos ^{-1}\left(\frac{2 t}{1+t^2}\right), y=\sec ^{-1}\left(\sqrt{1+t^2}\right)$
View full question & answer→Question 483 Marks
Find $\frac{d y}{d x}$ if$x=\left(t+\frac{1}{t}\right)^a, y=a^{t+\frac{1}{t}}$, where $a>0, a \neq 1$ and $t \neq 0$
View full question & answer→Question 493 Marks
Find $\frac{d y}{d x}$ if$x=\sqrt{a^2+m^2}, y=\log \left(a^2+m^2\right)$
View full question & answer→Question 503 Marks
If $y=x^{x^{x^{-\infty}}}$, then show that $\frac{d y}{d x}=\frac{y^2}{x(1-\log y)}$
View full question & answer→