Solve the following Question.(1 Marks)

Question 11 Mark

Find $\frac{d^2 y}{d x^2}$, if $y =\log x$

Answer

$
y=\log x
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x}
$
Differentiating again w.r.t. $x$, we get
$
\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{1}{x}\right)=-\frac{1}{x^2}
$

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Question 21 Mark

Find $\frac{d^2 y}{d x^2}$ if,
$
y=e^{\log x}
$

Answer

$
y=e^{\log x}=x
$
$
\text { ... }\left[\because a^{\log _a x}=x\right]
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{d}{d x}(x)=1
$
Differentiating again w.r.t. $x$, we get
$
\frac{d^2 y}{d x^2}=\frac{d}{d x}(1)=0
$

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Question 31 Mark

Find $\frac{d^2 y}{d x^2}$ if,
$
y = e ^{(2 x +1)}
$

Answer

$
y=e^{(2 x+1)}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{d}{d x}\left[e^{(2 x+1)}\right]=e^{(2 x+1)} \cdot \frac{d}{d x}(2 x+1) \\
& =e^{(2 x+1)} \times(2 \times 1+0)=2 e^{(2 x+1)}
\end{aligned}
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{d}{d x}\left[2 e^{(2 x+1)}\right]=2 \frac{d}{d x}\left[e^{(2 x+1)}\right] \\
& =2 e^{(2 x+1)} \cdot \frac{d}{d x}(2 x+1)=2 e^{(2 x+1)} \times(2 \times 1+0) \\
& =4 e^{(2 x+1)} .
\end{aligned}
$

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Question 41 Mark

Find $\frac{d^2 y}{d x^2}$ if,
$
y = e ^{ x }
$

Answer

$
y=e^x
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{d}{d x}\left(e^x\right)=e^x
$
Differentiating again w.r.t. $x$, we get
$
\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(e^x\right)=e^x
$

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Question 51 Mark

Find $\frac{d^2 y}{d x^2}$ if,
$
y=x^{-7}
$

Answer

$
y=x^{-7}
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{d}{d x}\left(x^{-7}\right)=-7 x^{-8}
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{d}{d x}\left(-7 x^{-8}\right)=-7 \frac{d}{d x}\left(x^{-8}\right) \\
& =(-7)(-8) x^{-9}=56 x^{-9} .
\end{aligned}
$

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Question 61 Mark

Find $\frac{d^2 y}{d x^2}$ if,
$
y=x^5
$

Answer

$
y=x^5
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{d}{d x}\left(x^5\right)=5 x^4
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{d}{d x}\left(5 x^4\right)=5 \frac{d}{d x}\left(x^4\right) \\
& =5 \times 4 x^3=20 x^3 .
\end{aligned}
$

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Question 71 Mark

Find $\frac{d^2 y}{d x^2}$ if,
$
y=\sqrt{ } x
$

Answer

$
y=\sqrt{x}
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{d}{d x}(\sqrt{x})=\frac{1}{2 \sqrt{x}}
$
Differentiating again w.r.t. $x$, we get
$
\begin{aligned}
\frac{d^2 y}{d x^2} & =\frac{d}{d x}\left(\frac{1}{2 \sqrt{x}}\right)=\frac{1}{2} \frac{d}{d x}\left(x^{-\frac{1}{2}}\right) \\
& =\frac{1}{2} \cdot\left(-\frac{1}{2}\right) x^{-\frac{1}{2}-1}=-\frac{1}{4} x^{-\frac{3}{2}}
\end{aligned}
$

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Question 81 Mark

Find $\frac{d y}{d x}$ if, :
Differentiate $5^x$ with respect to $\log x$.

Answer

Let $u=5^x$ and $v=\log x$
Then we want to find $\frac{d u}{d v}$
Differentiating $u$ and $v$ w.r.t. $x$, we get
$
\begin{gathered}
\frac{d u}{d x}=\frac{d}{d x}\left(5^x\right)=5^x \cdot \log 5 \\
\text { and } \frac{d v}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x} \\
\therefore \frac{d u}{d v}=\frac{(d u / d x)}{(d v / d x)}=\frac{5^x \cdot \log 5}{\left(\frac{1}{x}\right)}
\end{gathered}
$
$
=x \cdot 5^x \cdot \log 5 \text {. }
$

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Question 91 Mark

Find $\frac{d y}{d x}$ if, :
$
\sqrt{x}+\sqrt{y}=\sqrt{a}
$

Answer

$
\sqrt{x}+\sqrt{ } y=\sqrt{a}
$
Differentiating both sides w.r.t. $x_r$ we get
$
\begin{aligned}
& \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x}=0 \\
& \therefore \frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}}
\end{aligned}
$
$
\therefore \frac{d y}{d x}=-\frac{2 \sqrt{y}}{2 \sqrt{x}}=-\sqrt{\frac{y}{x}} \text {. }
$

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