Solve the Following Question.(2 Marks)

Question 12 Marks

If $\text{y}=\mid\text{x}-\text{x}^2\mid,$ then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$

Answer

We have
$\text{y}=\mid\text{x}-\text{x}^2\mid$
$\Rightarrow\text{y}=\mid\text{x}(1-\text{x})\mid$
$\Rightarrow\text{y}=\begin{cases}\text{x} -\text{x}^2 &\text{if}&0\leq\times\leq1\\\text{x}^2-\text{x} & \text{if}&\text{x}<0\ \text{or}\ \text{x}>1\end{cases}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\begin{cases}1-2\text{x}&\text{if}&0\leq\times\leq1\\2\text{x}-1&\text{if}&\text{x}\times0\ \text{or}\ \text{x}>1 \end{cases}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\begin{cases}-2&\text{if}&0\leq\times\leq1\\2&\text{if}&\text{x}<\ \text{or}\ \text{x}\geq\end{cases}$

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Question 22 Marks

Find $\frac{\text{d}^2\text{y}}{\text{dx}^2},$ where $\text{y}=\log\Big(\frac{\text{x}^2}{\text{e}^2}\Big)$

Answer

Here
$\text{y}=\log\Big(\frac{\text{x}^2}{\text{e}^2}\Big)$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\frac{\text{x}^2}{\text{e}^2}}\times\frac{2\text{x}}{\text{e}^2}=\frac{2}{\text{x}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-2}{\text{x}^2}$

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Question 32 Marks

Find the second order derivatives of the following functions:$\log(\sin\text{x})$

Answer

We have,
$\text{y}=\log(\sin\text{x})$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sin\text{x}}\times\cos\text{x}=\cot\text{x}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\text{cosec}^2\text{x}$

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Question 42 Marks

If $\text{y}=\mid\log_\text{e}\text{x}\mid $ find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$

Answer

Here,
$\text{y}=\mid\log_\text{e}\text{x}\mid$
$=\begin{cases}-\log_\text{e}\text{x}&\text{if}&0<\text{x}< 1\\\log_\text{e}\text{x}&\text{if}&\text{x}>1\end{cases}$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\begin{cases}\frac{-1}{\text{x}}&\text{if}&0<\text{x}<1\\\frac{1}{\text{x}}&\text{if}&\text{x}>1\end{cases}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\begin{cases}\frac{1}{\text{x}^2}&\text{if}&0<\text{x}<1\\\frac{-1}{\text{x}^2}&\text{if}&\text{x}>1 \end{cases}$

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Question 52 Marks

If $\text{x}=\text{f}(\text{t})$ and $\text{y}=\text{g}(\text{t}),$ then write the value of $\frac{\text{d}^2\text{y}}{\text{dx}^2}.$

Answer

We are given
$\text{x}=\text{f}(\text{t})$
$\text{Y}=\text{g}(\text{t})$
Then $\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\times\frac{\text{dt}}{\text{dx}}=\frac{\text{g}'(\text{t})}{\text{f}'(\text{t})}$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\frac{\text{f}'(\text{t})\text{g}'\text({t})-\text{g}'(\text{t})\text{f}''\text({t})}{[\text{f}'(\text{t})]^3}$

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Question 62 Marks

If $\text{x}=2\text{ at},\text{y}=\text{at}^2,$ where a is a constant, then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}\text{ at}\text{ x}=\frac{1}{2}.$

Answer

Here,
$\text{x}=2\text{at}\ \text{and}\ \text{y}=\text{at}^2$
Differentiating w.r.t.t, we get
$\frac{\text{dx}}{\text{dt}}=2\text{a}\ \text{and}\ \frac{\text{dy}}{\text{dt}}=2\text{at}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{2\text{at}}{2\text{a}}=\text{t}$
Differentiating w.r.t.t, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=1\times\frac{\text{dt}}{\text{dx}}=\frac{1}{2\text{a}}$
Now $\Big[\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big]_{\text{x}=\frac{1}{2}}=\frac{1}{2\text{a}}$

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Question 72 Marks

If $\text{x}=\text{t}^2$ and $\text{y}=\text{t}^3$ then find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$

Answer

Here,
$\text{x}=\text{t}^2\ \text{and}\ \text{y}=\text{t}^3$
$\Rightarrow\frac{\text{dx}}{\text{dt}}=2\text{t}\ \text{and}\ \frac{\text{dy}}{\text{dt}^2}=3\text{t}^2$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{3\text{t}}{2}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{3}{2}\frac{\text{dt}}{\text{dx}}=\frac{3}{4\text{t}}$

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Question 82 Marks

If $\text{y}=\cot\text{x}$ show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$

Answer

$\text{y}=\cot\text{x}$
Differentiating w.r.t.x
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-\text{cosec}^2\text{x}$
Differentiating w.r.t.x
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=-[2\text{cosec}\text{ x}(-\text{cosec}^2\times\cot\text{x})]=-2\frac{\text{dy}}{\text{dx}}.\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}+2\text{y}\frac{\text{dy}}{\text{dx}}=0$
Hence proved.

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Question 92 Marks

If $\text{y}=1-\text{x}+\frac{\text{x}^2}{2!}-\frac{\text{x}^3}{3!}+\frac{\text{x}^4}{4!}+...\infty$ then write $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ in terms of y.

Answer

Here,
$\text{y}=1-\text{x}+\frac{\text{x}^2}{2!}-\frac{\text{x}^3}{3!}+\frac{\text{x}^4}{4!}+...\infty$
Thus
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1+\frac{2\text{x}}{2!}-\frac{3\text{x}^2}{3!}+\frac{4\text{x}^3}{4!}...\infty$
$=-1+\text{x}-\frac{\text{x}^2}{2!}+\frac{\text{x}^3}{3!}...\infty$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=1-\frac{2\text{x}}{2!}+\frac{3\text{x}^2}{3!}-\frac{4\text{x}^3}{4!}+...\infty$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=1-\frac{2\text{x}}{2!}+\frac{3\text{x}^2}{3!}-\frac{4\text{x}^3}{4!}+...\infty$
$=1-\text{x}+\frac{\text{x}^2}{2!}-\frac{\text{x}^3}{3!}+...\infty$
$=\text{y}$

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Question 102 Marks

If $\text{y}=\text{x}+\text{e}^\text{x},$ find $\frac{\text{d}^2\text{x}}{\text{dy}^2}.$

Answer

Here,
$\text{y}=\text{x}+\text{e}^\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\text{e}^\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{e}^\text{x}}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\text{e}^\text{x}}{(1+\text{e}^\text{x})^2}$
$\frac{\text{dx}}{\text{dy}}=-\frac{-\text{e}^\text{x}}{(1+\text{e}^\text{x})^3}$

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Maths Solve the Following Question.(2 Marks)

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