datavorous/entrance-exam-dataset huggingface dataset:
If $x^{y}=e^{x-y}$ then $d y / d x$ is equal to which one of the following?

Options:
A $\frac{(x-y)}{(1+\log x)^{2}}$
B $\frac{y}{(1+\log x)}$
C $\frac{(x+y)}{(1+\log x)}$
D $\frac{(\log x)}{(1+\log x)^{2}}$

The correct answer is:
$\frac{(\log x)}{(1+\log x)^{2}}$

Solution:
$x^{y}=e^{x-y}$
Taking log both sides, we get $\Rightarrow y \cdot \log x=x-y$
$\Rightarrow y=\frac{x}{1+\log x}$
$\Rightarrow \frac{d y}{d x}=\frac{(1+\log x)-x\left(0+\frac{1}{x}\right)}{(1+\log x)^{2}}$
$=\frac{(1+\log x)-1}{(1+\log x)^{2}}=\frac{\log x}{(1+\log x)^{2}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 2)
If $x+y=t-\frac{1}{t}, x^{2}+y^{2}=t^{2}+\frac{1}{t^{2}}$, what is $\frac{d y}{d x}$ equal to?

Options:
A $\frac{1}{\mathrm{x}}$
B $-\frac{1}{x}$
C $\frac{1}{x^{2}}$
D $-\frac{1}{x^{2}}$

The correct answer is:
$\frac{1}{x^{2}}$

Solution:
Given that $x+y=t-\frac{1}{t}$ and $x^{2}+y^{2}=t^{2}+\frac{1}{t^{2}}$
$\therefore (\mathrm{x}+\mathrm{y})^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xy}$
$\Rightarrow\left(\mathrm{t}-\frac{1}{\mathrm{t}}\right)^{2}=\left(\mathrm{t}^{2}+\frac{1}{\mathrm{t}^{2}}\right)+2 \mathrm{xy}$
$-2=2 x y \Rightarrow x y=-1$
$(x-y)^{2}=(x+y)^{2}-4 x y$
$=\left(t-\frac{1}{t}\right)^{2}-4 \times-1=t^{2}+\frac{1}{t^{2}}-2+4=\left(t+\frac{1}{t}\right)^{2}$
$\mathrm{x}-\mathrm{y}=\mathrm{t}+\frac{1}{\mathrm{t}}$
$\Rightarrow \mathrm{x}=\mathrm{t}, \mathrm{y}=-\frac{1}{\mathrm{t}}$
$\mathrm{xy}=-1$
$\Rightarrow x \frac{d y}{d x}+y=0$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}=\frac{1}{\mathrm{t}^{2}}=\frac{1}{\mathrm{x}^{2}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2006 (Phase 1)
If \( x y z \) are all different and not equal to zero and \( \left|\begin{array}{cccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0 \)
then the value of \( x^{-1}+y^{-1}+z^{-1} \) is equal to

Options:
A \( x y z \)
B \( x^{-1} y^{-1} z^{-1} \)
C \( -x-y-z \)
D \( -1 \)

The correct answer is:
\( -1 \)

Solution:
Given that,
\[
\begin{array}{l}
\left|\begin{array}{ccc}
1+x & 1 & 1 \\
1 & 1+y & 1 \\
1 & 1 & 1+z
\end{array}\right|=0 \\
R_{1} \rightarrow R_{1}-R_{2} \text { and } R_{2} \rightarrow R_{2}-R_{3} \\
\left|\begin{array}{ccc}
x & -y & 0 \\
0 & y & -z \\
1 & 1 & 1+z
\end{array}\right|=0 \\
\Rightarrow x[y(1+z)+z]+y(z)=0 \\
\Rightarrow x y+y z+z x+x y z=0
\end{array}
\]
Divide by xyz both side, we get
\[
\begin{array}{l}
\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+1=0 \\
\Rightarrow x^{-1}+y^{-1}+z^{-1}=-1
\end{array}
\]

Topic: Mathematics, Determinants, KCET, KCET 2016
If \( x y z \) are not equal and \( \neq 0, \neq 1 \) the value of \( \left|\begin{array}{ccc}\log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z\end{array}\right| \) is equal to

Options:
A \( \log (x y z) \)
B \( \log (6 \times y z) \)
C \( 00 \)
D \( \log (x+y+z) \)

The correct answer is:
\( 00 \)

Solution:
Given that, $x, y, z \neq 0 \neq 1$
As we know, $\log 2 x-\log x=\log 2$
and $\log 3 x-\log x=\log 3$
Given that $\left|\begin{array}{ccc}\log x & \log y & \log z \\ \log 2 x & \log 2 y & \log 2 z \\ \log 3 x & \log 3 y & \log 3 z\end{array}\right|$
$R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$
$\mid \begin{array}{ll}\log x & \log y & \log z \\ \log 2 & \log 2 & \log 2 \\ \log 3 & \log 3 & \log 3 & \\ = & \log 2 \log 3 \\ \text { When the two rows of determinants are identical the value of determinant is equal to } 0 & \begin{array}{ccc}\log x & \log y & \log z \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{array} \mid\end{array}$

Topic: Mathematics, Determinants, KCET, KCET 2016
If \(x=\alpha, y=\beta, z=\gamma\) is the solution, for the system of equations
\(\begin{aligned}
2 x-y+8 z & =13 \\
3 x+4 y+5 z & =18 \\
5 x-2 y+7 z & =20
\end{aligned}\)
then \(\alpha \beta+\beta \gamma+\gamma \alpha=\)

Options:
A 1
B 0
C 7
D -3

The correct answer is:
7

Solution:
According to Cramer's rule,
\(\begin{aligned}
& x=\alpha=\frac{\Delta_1}{\Delta}, y=\beta=\frac{\Delta_2}{\Delta} \text { and } z=\gamma=\frac{\Delta_3}{\Delta} \\
& \text {where } \begin{aligned}
\Delta & =\left|\begin{array}{ccc}
2 & -1 & 8 \\
3 & 4 & 5 \\
5 & -2 & 7
\end{array}\right| \\
& =2(28+10)+1(21-25)+8(-6-20) \\
& =76-4-208=76-212=-136
\end{aligned}
\end{aligned}\)
\(\begin{aligned}
\Delta_1 & =\left|\begin{array}{ccc}
13 & -1 & 8 \\
18 & 4 & 5 \\
20 & -2 & 7
\end{array}\right| \\
& =13(28+10)+1(126-100)+8(-36-80) \\
& =(13 \times 38)+26-(8 \times 116) \\
& =494+26-928=-408 \\
\Delta_2 & =\left|\begin{array}{lll}
2 & 13 & 8 \\
3 & 18 & 5 \\
5 & 20 & 7
\end{array}\right| \\
& =2(126-100)-13(21-25)+8(60-90) \\
& =(2 \times 26)+(13 \times 4)-(8 \times 30) \\
& =52+52-240=-136 \\
\Delta_3 & =\left|\begin{array}{ccc}
2 & -1 & 13 \\
3 & 4 & 18 \\
5 & -2 & 20
\end{array}\right| \\
& =2(80+36)+1(60-90)+13(-6-20) \\
& =(2 \times 116)-30-(13 \times 26) \\
& =232-30-338=-136
\end{aligned}\)
So, \(\alpha=3, \beta=1, \gamma=1\)
\(\therefore \quad \alpha \beta+\beta \gamma+\gamma \alpha=3+1+3=7\)
Hence, option (c) is correct.

Topic: Mathematics, Determinants, AP EAMCET, AP EAMCET 2019 (23 Apr Shift 1)
If $\mathrm{xdy}=\mathrm{y} \mathrm{dx}+\mathrm{y}^{2} \mathrm{dy}, \mathrm{y}>0$ and $\mathrm{y}(1)=1$, then what is
$\mathrm{y}(-3)$ equal to?

Options:
A 3 only
B $-1$ only
C Both $-1$ and 3
D Neither $-1$ nor 3

The correct answer is:
3 only

Solution:
Given, $x d y=y d x+y^{2} d y$
$\Rightarrow 1=\frac{y}{x} \frac{d x}{d y}+\frac{y^{2}}{x}$
$\Rightarrow \frac{d x}{d y}+y=\frac{x}{y}$
$\Rightarrow \frac{d x}{d y}-\frac{x}{y}=-y$
$P=-\frac{1}{y}, Q=-y$
$\mathrm{IF}=e^{\int P d y}=e^{\int-\frac{1}{y} d y}=e^{-\log y}=\frac{1}{y}$
Now, Solution of d.E. $\mathrm{x}(\mathrm{I.F})=\int(\mathrm{Q} . \mathrm{I.F}) \mathrm{dy}$
$\frac{x}{y}=\int \frac{1}{y}(-y) d y+C$
$\Rightarrow \frac{x}{y}=\int-1 d y+C$
$\Rightarrow \frac{x}{y}=-y+C$
$y(1)=1$
$\frac{1}{1}=-1+C \Rightarrow C=2$
$\Rightarrow \frac{x}{y}=-y+2 \Rightarrow x=-y^{2}+2 y$
$\Rightarrow y(-3) \Rightarrow-3=-y^{2}+2 y$
$\Rightarrow y^{2}-2 y-3=0$
$\Rightarrow y=\frac{+2 \pm \sqrt{4+12}}{2}=\frac{2 \pm 4}{2}$
$\Rightarrow y=3,-1$
Since $y>0$ so $y=3$.

Topic: Mathematics, Differential Equations, NDA, NDA 2016 (Phase 2)
If $y=\left(1+x^{\frac{1}{4}}\right)\left(1+x^{\frac{1}{2}}\right)\left(1-x^{\frac{1}{4}}\right)$, then what is $\frac{d y}{d x}$ equal to?

Options:
A 1
B $-1$
C $x$
D $x^{\frac{1}{2}}$

The correct answer is:
$-1$

Solution:
Let $y=\left(1+x^{1 / 4}\right)\left(1+x^{1 / 2}\right)\left(1-x^{1 / 4}\right)$
$=\left(1+x^{1 / 4}\right)\left(1-x^{1 / 4}\right)\left(1+x^{1 / 2}\right)$
$=\left(1-x^{1 / 2}\right)\left(1+x^{1 / 2}\right)$
$\left(\because(a+b)(a-b)=a^{2}-b^{2}\right)$
$\begin{array}{l}
=(1-x)\left(\because(a+b)(a-b)=a^{2}-b^{2}\right) \\
\Rightarrow y=1-x
\end{array}$
Differentiate both side w.r.t ' $x$, $\frac{d y}{d x}=-1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 2)
If $\mathrm{y}^{2}=\mathrm{p}(\mathrm{x})$ is a polynomial of degree 3, then what is
$2 \frac{\mathrm{d}}{\mathrm{dx}}\left[\mathrm{y}^{3} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right]$ equal to ?

Options:
A $\mathrm{p}^{\prime}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})$
B $\mathrm{p}^{\prime \prime}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})$
C $\mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime \prime}(\mathrm{x})$
D $\begin{array}{ll}\text { A constant } & {}\end{array}$

The correct answer is:
$\mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime \prime}(\mathrm{x})$

Solution:
Given that $y^{2}=p(x)$
Differentiating
$\Rightarrow 2 \mathrm{yy}_{1}=\mathrm{p}^{\prime}(\mathrm{x})$
$\left[\right.$ here $\left.\mathrm{y}_{1}=\frac{\mathrm{dy}}{\mathrm{dx}}\right]$
$\Rightarrow \quad 2 \mathrm{y}_{1}=\frac{\mathrm{p}^{\prime}(\mathrm{x})}{\mathrm{y}}$
Differentiating again, $\Rightarrow \quad 2 \mathrm{y}_{2}=\frac{\mathrm{yp}^{\prime \prime}(\mathrm{x})-\mathrm{p}^{\prime}(\mathrm{x}) \mathrm{y}_{1}}{\mathrm{y}^{2}},\left[\mathrm{y}_{2}=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}\right]$
$\Rightarrow 2 \mathrm{y}_{2}=\frac{\mathrm{yp}^{\prime \prime}(\mathrm{x})-\frac{\mathrm{p}^{\prime}(\mathrm{x}) \cdot \mathrm{p}^{\prime}(\mathrm{x})}{2 \mathrm{y}}}{\mathrm{y}^{2}}$
$$
\begin{array}{l}
\quad=\frac{\left.2 \mathrm{y}^{2} \mathrm{p}^{\prime \prime}(\mathrm{x})-\mathrm{p}^{\prime}(\mathrm{x})\right)^{2}}{2 \mathrm{y}^{3}} \\
\Rightarrow \quad 2 \mathrm{y}^{3} \mathrm{y}_{2}=\frac{1}{2}\left[2 \mathrm{y}^{2} \mathrm{p}^{\prime \prime}(\mathrm{x})-\left(\mathrm{p}^{\prime}(\mathrm{x})\right)^{2}\right] \\
\Rightarrow \quad 2 \mathrm{y}^{3} \mathrm{y}_{2}=\frac{1}{2}\left[2 \mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})-\left(\mathrm{p}^{\prime}(\mathrm{x})\right)^{2}\right] \\
\Rightarrow \quad 2 \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{y}^{3} \mathrm{y}_{2}\right) \\
=\frac{1}{2}\left[2 \mathrm{p}^{\prime}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})+2 \mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})-2 \mathrm{p}^{\prime}(\mathrm{x}) \mathrm{p}^{\prime \prime}(\mathrm{x})\right] \\
=\mathrm{p}(\mathrm{x}) \mathrm{p}^{\prime \prime \prime}(\mathrm{x})
\end{array}
$$

Topic: Mathematics, Differential Equations, NDA, NDA 2006 (Phase 1)
If $y=a \cos 2 x+b \sin 2 x$, then

Options:
A $\frac{d^{2} y}{d x^{2}}+y=0$
B $\frac{d^{2} y}{d x^{2}}+2 y=0$
C $\frac{d^{2} y}{d x^{2}}-4 y=0$
D $\frac{d^{2} y}{d x^{2}}+4 y=0$

The correct answer is:
$\frac{d^{2} y}{d x^{2}}+4 y=0$

Solution:
The equation of circle touching $\mathrm{y}$ -axis at origin is $(\mathrm{x}-\alpha)^{2}+(\mathrm{y}-0)^{2}=\alpha^{2}$
$\Rightarrow \mathrm{x}^{2}+\alpha^{2}-2 \alpha \mathrm{x}+\mathrm{y}^{2}=\alpha^{2}$
$\Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}-2 \alpha \mathrm{x}=0$
$\Rightarrow x+\frac{y^{2}}{x}-2 \alpha=0$
Differentiating, $1+\frac{x \cdot 2 y \cdot \frac{d y}{d x}-y^{2}}{x^{2}}=0$
$\Rightarrow x^{2}+2 x y \cdot \frac{d y}{d x}-y^{2}=0$

Topic: Mathematics, Differential Equations, NDA, NDA 2019 (Phase 1)
If $\mathrm{y}$ -axis touches the circle
$x^{2}+y^{2}+g x+f y+\frac{c}{4}=0$, then the normal at this point intersects the circle at the point

Options:
A $\left(-\frac{\mathrm{g}}{2},-\frac{\mathrm{f}}{2}\right)$
B $\left(-\mathrm{g},-\frac{\mathrm{f}}{2}\right)$
C $\left(-\frac{\mathrm{g}}{2}, \mathrm{f}\right)$
D $(-\mathrm{g},-\mathrm{f})$

The correct answer is:
$\left(-\mathrm{g},-\frac{\mathrm{f}}{2}\right)$

Solution:

Topic: Mathematics, Circle, NDA, NDA 2018 (Phase 2)
If $\mathrm{y}=\cos ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

Options:
A $-\frac{2}{1+x^{2}}$ for all $|x| < 1$
B $-\frac{2}{1+x^{2}}$ for all $|x|>1$
C $\frac{2}{1+\mathrm{x}^{2}}$ for all $|\mathrm{x}| < 1$
D None of the above

The correct answer is:
$-\frac{2}{1+x^{2}}$ for all $|x| < 1$

Solution:
$\mathrm{y}=\cos ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right)$
Put $\mathrm{x}=\tan \theta \Rightarrow \theta=\tan ^{-1} \mathrm{x}$
$\mathrm{y}=\cos ^{-1}\left(\frac{2 \tan \theta}{\left(1+\tan ^{2} \theta\right)}\right)=\cos ^{-1}(\sin 2 \theta)$
$=\cos ^{-1}\left(\cos \left(\frac{\pi}{2}-2 \theta\right)\right)=\frac{\pi}{2}-2 \theta .$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{2}-2 \theta\right)$
$=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\pi}{2}-2 \tan ^{-1} \mathrm{x}\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2}{1+\mathrm{x}^{2}}$, when $|\mathrm{x}| < 1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
If $y=\cos t$ and $x=\sin t$, then what is $\frac{d y}{d x}$ equal to?

Options:
A $\mathrm{xy}$
B $\mathrm{x} / \mathrm{y}$
C $-\mathrm{y} / \mathrm{x}$
D $-x / y$

The correct answer is:
$-x / y$

Solution:
Let $y=\cos t, x=\sin t$
$\frac{\mathrm{dy}}{\mathrm{dt}}=-\sin \mathrm{t}, \frac{\mathrm{dx}}{\mathrm{dt}}=\cos \mathrm{t}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}=-\frac{\sin \mathrm{t}}{\cos \mathrm{t}}=-\frac{\mathrm{x}}{\mathrm{y}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 1)
If $\mathrm{y}=(\cos \mathrm{x})^{(\cos \mathrm{x})^{(\cos \mathrm{x})^{...\infty}}}$, then $\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

Options:
A $-\frac{y^{2} \tan x}{1-y \operatorname{In}(\cos x)}$
B $\frac{y^{2} \tan x}{1+y \operatorname{In}(\cos x)}$
C $\frac{y^{2} \tan x}{1-y \ln (\sin x)}$
D $\frac{y^{2} \sin x}{1+y \operatorname{In}(\sin x)}$

The correct answer is:
$-\frac{y^{2} \tan x}{1-y \operatorname{In}(\cos x)}$

Solution:
$\mathrm{y}=(\cos \mathrm{x})^{(\cos \mathrm{x})^{(\cos \mathrm{x})^{...\infty}}}$
$\Rightarrow \mathrm{y}=(\cos \mathrm{x})^{\mathrm{y}}$
$\Rightarrow \log \mathrm{y}=\mathrm{y} \cdot \log \cos \mathrm{x}$
Differentiating on both sides, we get
$\frac{1}{y} \cdot \frac{d y}{d x}=y(-\tan x)+\log \cos x \cdot \frac{d y}{d x}$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\left(\frac{1}{\mathrm{y}}-\log \cos \mathrm{x}\right)=-\mathrm{y} \tan \mathrm{x}$
$\Rightarrow \frac{d y}{d x}=\frac{-y \tan x}{\frac{1}{y}-\log \cos x}=\frac{-y^{2} \tan x}{1-y \log \cos x}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
If $y=\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$, where $0 < x < \frac{\pi}{2}$, then
$\frac{\mathrm{dy}}{\mathrm{dx}}$ is equal to

Options:
A $\frac{1}{2}$
B 2
C $\sin x+\cos x$
D $\sin x-\cos x$

The correct answer is:
$\frac{1}{2}$

Solution:
$\mathrm{y}=\cot ^{-1}\left[\frac{\sqrt{1+\sin \mathrm{x}}+\sqrt{1-\sin \mathrm{x}}}{\sqrt{1+\sin \mathrm{x}}-\sqrt{1-\sin \mathrm{x}}}\right]$
$\mathrm{y}=\cot ^{-1}$
$\left[\begin{array}{l}\sqrt{\cos ^{2} \frac{\mathrm{x}}{2}+\sin ^{2} \frac{\mathrm{x}}{2}+2 \sin \frac{\mathrm{x}}{2} \cos \frac{\mathrm{x}}{2}} \\ +\sqrt{\cos ^{2} \frac{\mathrm{x}}{2}+\sin ^{2} \frac{\mathrm{x}}{2}-2 \sin \frac{\mathrm{x}}{2} \cos \frac{\mathrm{x}}{2}}{\sqrt{\cos ^{2} \frac{\mathrm{x}}{2}+\sin ^{2} \frac{\mathrm{x}}{2}+2 \sin \frac{\mathrm{x}}{2} \cos \frac{\mathrm{x}}{2}}} \\ -\sqrt{\cos ^{2} \frac{\mathrm{x}}{2}+\sin ^{2} \frac{\mathrm{x}}{2}-2 \sin \frac{\mathrm{x}}{2} \cos \frac{\mathrm{x}}{2}}\end{array}\right]$
$y=\cot ^{-1}\left[\frac{\sqrt{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}+\sqrt{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}}{\sqrt{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}-\sqrt{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}}\right]$
$y=\cot ^{-1}\left[\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}+\sin \frac{x}{2}}\right]$
$y=\cot ^{-1}\left[\frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}}\right]=\cot ^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2}$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2015 (Phase 2)
If $\mathrm{y}=\mathrm{e}^{\mathrm{x}^{2}} \sin 2 \mathrm{x}$, then what is $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=\pi$ equal to?

Options:
A $(1+\pi) \mathrm{e}^{\pi^{2}}$
B $2 \pi \mathrm{e}^{\pi^{2}}$
C $2 \mathrm{e}^{\pi^{2}}$
D $\mathrm{e}^{\pi^{2}}$

The correct answer is:
$2 \mathrm{e}^{\pi^{2}}$

Solution:
$y=e^{x^{2}} \cdot \sin 2 x$
$\frac{d y}{d x}=2 \cdot e^{x^{2}} \cdot \cos 2 x+2 x e^{x^{2}} \cdot \sin 2 x$
$=2 e^{x^{2}}(\cos 2 x+x \sin 2 x)$
$\left.\frac{d y}{d x}\right|_{x=\pi}=2 e^{\pi^{2}}(\cos 2 \pi+\pi \cdot \sin 2 \pi)$
$=2 e^{\pi^{2}}(1+0)$
$=2 \cdot e^{\pi^{2}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2018 (Phase 1)
If $\mathrm{y}=\mathrm{f}(\mathrm{x}), \mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}$ and $\mathrm{q}=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$, then what is $\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}$ equal
to?

Options:
A $-\frac{\mathrm{q}}{\mathrm{p}^{2}}$
B $-\frac{\mathrm{q}}{\mathrm{p}^{3}}$
C $\frac{1}{\mathrm{q}}$
D $\frac{\mathrm{q}}{\mathrm{p}^{2}}$

The correct answer is:
$-\frac{\mathrm{q}}{\mathrm{p}^{3}}$

Solution:
As given, $y=f(x), p=\frac{d y}{d x}$ and $q=\frac{d^{2} y}{d x^{2}}$
$\frac{\mathrm{dx}}{\mathrm{dy}}=\frac{1}{\mathrm{p}} \Rightarrow \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dy}^{2}}=\frac{-1}{\mathrm{p}^{2}} \cdot \frac{\mathrm{dp}}{\mathrm{dy}}$
$\frac{\mathrm{dp}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{q}$
$\frac{d p}{d y}=\frac{d p}{d x} \cdot \frac{d x}{d y}=q \cdot \frac{1}{p}=\frac{q}{p}$
$\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{d} \mathrm{y}^{2}}=\Rightarrow-\frac{1}{\mathrm{p}^{2}} \times \frac{\mathrm{q}}{\mathrm{p}}=\frac{-\mathrm{q}}{\mathrm{p}^{3}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 1)
If $\mathrm{y}=\frac{1}{\log _{10} \mathrm{x}}$, then what is $\frac{\mathrm{dy}}{\mathrm{dx}}$ equal to?

Options:
A $x$
B $\mathrm{x} \log _{\mathrm{e}} 10$
C $-\frac{\left(\log _{x} 10\right)^{2}\left(\log _{10} e\right)}{x}$
D $x \log _{10} e$

The correct answer is:
$-\frac{\left(\log _{x} 10\right)^{2}\left(\log _{10} e\right)}{x}$

Solution:
Differentiating the given function, $\mathrm{y}=\frac{1}{\log _{10} \mathrm{x}}$
We get, $\frac{d y}{d x}=-\frac{1}{\left(\log _{10} x\right)^{2}} \cdot \frac{1}{x} \log _{10} e$
$\Rightarrow \frac{d y}{d x}=-\frac{\left(\log _{x} 10\right)^{2} \cdot \log _{10} e}{x}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 1)
If $\mathrm{y}=\log _{10} \mathrm{x}+\log _{\mathrm{x}} 10+\log _{\mathrm{x}} \mathrm{x}+\log _{10} 10$ then what is
$\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}=10}$ equal to?

Options:
A 10
B 2
C 1
D 0

The correct answer is:
0

Solution:
$y=\log _{10} x+\log _{x} 10+\log _{x} x+\log _{10} 10$
$y=\log _{10} x+\log _{x} 10+1+1$
Differentiating equation w.r.t. $\mathrm{x}$
$\frac{d y}{d x}=\frac{1}{x \log _{e} 10}-\frac{1}{\left(\log _{10} x\right)^{2}} \cdot \frac{1}{(x \log 10)}$
$=\frac{1}{x \log _{e} 10}\left[1-\frac{1}{\left(\log _{10} x\right)^{2}}\right]$
$\left(\frac{d y}{d x}\right)_{x=10}=\frac{1}{10 \log _{e} 10}[1-1]=0$
$\left[\begin{array}{l}\text { Note: } \log _{x} 10=\frac{\log _{10} 10}{\log _{10} x}=\frac{1}{\log _{10} x} \\ \frac{d}{d x}\left[\frac{1}{\log _{10} x}\right]=-\left(\log _{10} x\right)^{-2} \times \frac{1}{x \log _{e} 10} \\ =-\frac{1}{\left(\log _{10} x\right)^{2} x \log _{e} 10}\end{array}\right]$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2016 (Phase 1)
If $y=\ell n\left(e^{m x}+e^{-m x}\right)$, then what is $\frac{d y}{d x}$ at $x=0$ equal to?

Options:
A -1
B 0
C 1
D 2

The correct answer is:
0

Solution:
$\mathrm{y}=\ln (\mathrm{emx}+\mathrm{e}-\mathrm{mx})$
$\mathbf{p}$ $\quad \frac{d y}{d x}=\frac{1}{e^{m x}+e^{-m x}} \cdot \frac{d}{d x}(\mathrm{emx}+\mathrm{e}-\mathrm{mx})$
$=\frac{m e^{m x}-m e^{-m x}}{e^{m n}+e^{-m x}}=\frac{m\left(e^{m x}-e^{-m x}\right)}{e^{m x}+e^{-m x}}$
$=\frac{m\left[e^{m n}-\frac{1}{e^{m x}}\right]}{e^{m x}+\frac{1}{x^{n x}}}=\frac{m\left[e^{2 m x}-1\right]}{e^{2 m x}+1}$
so, $\left.\frac{d y}{d x}\right|_{x=0}=\frac{m\left(e^{0}-1\right)}{e^{0}+1}=m(0)=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 2)
If $y=\ell \mathrm{n} \sqrt{\tan x}$, then what is the value of $\frac{d y}{d x}$ at $x=\frac{\pi}{4}$?

Options:
A 0
B -1
C $1 / 2$
D 1

The correct answer is:
1

Solution:
Let $\mathrm{y}=\ln \sqrt{\tan x}$
Differentiate both side w.r.t $^{\prime} x^{\prime}$
$\frac{d y}{d x}=\frac{1}{\sqrt{\tan x}} \cdot \frac{1}{2 \sqrt{\tan x}} \cdot \sec ^{2} x$
Now, $\frac{d y}{d x}$ at $x=\pi / 4$
$=\frac{1}{\sqrt{\tan \frac{\pi}{4}}} \times \frac{1}{2 \sqrt{\tan \frac{\pi}{4}}} \times \frac{1}{\cos ^{2}(\pi / 4)}$
$=\frac{1}{2} \times 1 \times \frac{1}{\left(\frac{1}{\sqrt{2}}\right)^{2}}=\frac{1}{2} \times 1 \times 2=1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 2)
If $y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right)$, then $\frac{d y}{d x}$ is equal to

Options:
A 0
B 1
C $\frac{x-1}{x+1}$
D $\frac{x+1}{x-1}$

The correct answer is:
0

Solution:
$\mathrm{y}=\sec ^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)+\sin ^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)$
$=\cos ^{-1}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)+\sin ^{-1}\left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)$
$=\frac{\pi}{2}\left(\because \sin ^{-1} \mathrm{x}+\cos ^{-1} \mathrm{x}=\frac{\pi}{2}\right) \quad \therefore \frac{\mathrm{dy}}{\mathrm{dx}}=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2017 (Phase 2)
If $y=\sec ^{2} \theta+\cos ^{2} \theta$, where $0 < \theta < \frac{\pi}{2}$, then which one of
the following is correct?

Options:
A $y=0$
B $0 \leq y \leq 2$
C $\mathrm{y} \geq 2$
D None of these

The correct answer is:
$\mathrm{y} \geq 2$

Solution:
Since, $\cos ^{2} \theta$ lies between 0 and 1 therefore,
$\sec ^{2} \hat{\theta}+\cos ^{2} \hat{\theta}>2, \forall 0 < \dot{\theta} < \frac{\pi}{2}$
$\therefore \quad y \geq 2$

Topic: Mathematics, Trigonometric Equations, NDA, NDA 2010 (Phase 2)
If $y=\sin ^{-1}\left(\frac{4 x}{1+4 x^{2}}\right)$, then what is $\frac{d y}{d x}$ equal to?

Options:
A $\frac{1}{1+4 x^{2}}$
B $-\frac{1}{1+4 x^{2}} \quad[2010-I]$
C $\frac{4}{1+4 x^{2}}$
D $\frac{4 x}{1+4 x^{2}}$

The correct answer is:
$\frac{4}{1+4 x^{2}}$

Solution:
Let $y=\sin ^{-1}\left(\frac{4 x}{1+4 x^{2}}\right)=\sin ^{-1}\left(\frac{2 \cdot 2 x}{1+(2 x)^{2}}\right)$
Put $2 x=\tan \theta \Rightarrow \theta=\tan ^{-1} 2 x$
$\therefore y=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$
$=\sin ^{-1}(\sin 2 \theta)=2 \theta\left(\because \sin 2 \theta=\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$
$=2 \tan ^{-1} 2 x$
On differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{2}{1+(2 x)^{2}} \cdot 2=\frac{4}{1+4 x^{2}}$
ALTERNATE SOLUTION
$y=\sin ^{-1}\left(\frac{4 x}{1+4 x^{2}}\right)$
$\frac{d y}{d x}=\frac{1}{\sqrt{1-\left(\frac{4 x}{1+4 x^{2}}\right)^{2}} \times \frac{\left(1+4 x^{2}\right) 4-4 x(8 x)}{\left(1+4 x^{2}\right)^{2}}}$
$=\frac{1+4 x^{2}}{\sqrt{\left(1+4 x^{2}\right)^{2}-16 x^{2}}} \times \frac{4+16 x^{2}-32 x^{2}}{\left(1+4 x^{2}\right)^{2}}$
$=\frac{4-16 x^{2}}{\left(1+4 x^{2}\right) \sqrt{1-8 x^{2}+16 x^{4}}}$
$=\frac{4-16 x^{2}}{\left(1+4 x^{2}\right)\left(1-4 x^{2}\right)}$
$=\frac{(2)^{2}-(4 x)^{2}}{\left(1+4 x^{2}\right)(1-2 x)(1+2 x)}=\frac{(2+4 x) 2}{\left(1+4 x^{2}\right)(1+2 x)}=\frac{4}{1+4 x^{2}}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 1)
If $y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}$, what is $\frac{d y}{d x}$ equal to?

Options:
A $\cos ^{-1} x+\cos ^{-1} \sqrt{1-x^{2}}$
B $\frac{1}{\cos x}+\frac{1}{\cos \sqrt{1-x^{2}}}$
C $\frac{\pi}{2}$
D $0$

The correct answer is:
$0$

Solution:
Given function is:
$y=\sin ^{-1} x+\sin ^{-1} \sqrt{1-x^{2}}$
On differentiating, w.r.t. $\mathrm{x}$, we get
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}+\frac{1}{\sqrt{1-1+\mathrm{x}^{2}}} \cdot \frac{1}{2 \sqrt{1-\mathrm{x}^{2}}}(-2 \mathrm{x})$
$=\frac{1}{\sqrt{1-\mathrm{x}^{2}}}-\frac{1}{\sqrt{1-\mathrm{x}^{2}}}=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 1)
If $y=\sin (a x+b)$, then what is $\frac{d^{2} y}{d x^{2}}$ at $x=-\frac{b}{a}$, where $a, b$ are constants and $\mathrm{a} \neq 0$?

Options:
A 0
B $-1$
C $\sin (a-b)$
D $\sin (a+b)$

The correct answer is:
0

Solution:
Let $y=\sin (a x+b)$
$\Rightarrow \frac{d y}{d x}=a \cos (a x+b)$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=-a^{2} \sin (a x+b)$
Now, $\frac{d^{2} y}{d x^{2}}$ at $x=-\frac{b}{a}$ is
$-a^{2} \sin \left(a\left(-\frac{b}{a}\right)+b\right)=-a^{2} \sin 0=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 1)