Solve the Following Question.(4 Marks)

Question 14 Marks

If $y=f(u)$ is a differentiable function of $u$ and $u=g(x)$ is a differentiable function of $x$ such that the composite function $y=f[g(x)]$ is a differentlable function of $x$ then prove that
$\frac{d y}{d x}=\frac{d y}{d u} \times \frac{d u}{d x}$
Hence find $\frac{d y}{d x}$ if $y=\sqrt{x^2+5}$

Answer

Given: $y=f(u)$ and $u=g(x)$, where $u$ is not a constant function.
Let $\delta x$ the small increment in $x$ and let $\delta u$ and $\delta y$ be the corresponding small increments in $u$ and $y$ respectively.
$\therefore \delta x \neq 0, \delta u \neq 0$ and $\delta y \neq 0$.
We have : $\frac{\delta y}{\delta x}=\frac{\delta y}{\delta u} \cdot \frac{\delta u}{\delta x}$
Taking limit on both sides $\delta x \rightarrow 0$.
$\lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta x}\right)=\lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta u}\right) \times \lim _{\delta x \rightarrow u}\left(\frac{\delta u}{\delta x}\right)$
As $\delta x=0$ we get $\delta u \rightarrow 0 \quad$ (as $u$ is a continuous function of $x$ )
$\therefore \lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta x}\right)=\lim _{\delta u \rightarrow 0}\left(\frac{\delta y}{\delta u}\right) \times \lim _{\delta x \rightarrow 0}\left(\frac{\delta u}{\delta x}\right)....(1)$
Sine $y$ is a differentiable function of $u$ and $u$ is a differentiable function of $x$,
we get $\lim _{\delta u \rightarrow 0}\left(\frac{\delta y}{\delta u}\right)=\frac{d y}{d u}$ and $\lim _{\overline{\delta x} \rightarrow x}\left(\frac{\delta u}{\delta x}\right)=\frac{d u}{d x}....(2)$
From $(1)$ and $(2)$
$\lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta x}\right)=\frac{d y}{d u} \cdot \frac{d u}{d x}....(3)$
The $\text{R.H.S}$. of $(3)$ exists and is finite, it implies that $\text{L.H.S.}$ of $(3)$ also exists and is finite.
$\therefore \lim _{\delta x \rightarrow 0}\left(\frac{\delta y}{\delta x}\right)=\frac{d y}{d x}$
$\therefore$ Equation $(i)$ becomes
$\therefore \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}$
Now, $y=\sqrt{x^2+5}$
Let $u=x^2+5$
$\therefore \frac{d u}{d x}=2 x$
Also $y=\sqrt{u}$
$\therefore \frac{d y}{d u}=\frac{1}{2 \sqrt{u}}=\frac{1}{2 \sqrt{x^2+5}}$
$\therefore \frac{d y}{d x}=\frac{d y}{d u} \cdot \frac{d u}{d x}$
$=\frac{1}{2 \sqrt{x^2+5}}(2 x)$
$=\frac{x}{\sqrt{x^2+5}}$

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Question 24 Marks

If $x=f(t), y=g(t)$ are differentiable functions of parameter ' $t$ ' then prove that $y$ is a differentiable function of ' $x$ ' and $\frac{d y}{d x}=\frac{\left(\frac{d y}{d t}\right)}{\left(\frac{d x}{d t}\right)}, \frac{d x}{d t} \neq 0$. Hence find $\frac{d y}{d x}$ if $x=a \cos t, y=a \sin t$.

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Question 34 Marks

If $\sqrt{1-x^2}+\sqrt{1-y^2}=a(x-y)$, show that $\frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}}$

Answer

$\sqrt{1-x^2}+\sqrt{1-y^2}= a ( x - y )$
Put $x = \sin \theta , y = \sin \Phi$
$\therefore \theta = \sin^{−1} x, \Phi = \sin^{−1} y$

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Question 44 Marks

If $x=f(t)$ and $y=g(t)$ are differentiable function of $t$ so that $y$ is a differentiable function of $x$ and $\frac{d x}{d t} \neq 0$, then prove that :
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Hence find $\frac{d y}{d x}$ if $x=\sin t$ and $y=\operatorname{cost} t$.

Answer

Given: $x=f(t), y=g(t)$
Let $\delta t$ be a small increment in $t, \delta x, \delta y$ be. the corresponding increments in $x$ and $y$ respectively.
Consider
$\frac{\delta y}{\delta x}=\frac{\frac{\delta y}{\delta t}}{\frac{\delta x}{\delta t}}, \frac{\delta x}{\delta t} \neq 0$
Taking the limit as $\delta t \rightarrow 0$,
$\lim _{x \rightarrow 00} \frac{\delta y}{\delta x}=\lim _{x t \rightarrow 0} \frac{\delta y}{\frac{\delta t}{\delta x}}, \lim _{x \rightarrow 00} \frac{\delta x}{\delta t} \neq 0$
As $\delta t \rightarrow 0, \delta x \rightarrow 0$
$\therefore \lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta t}}{\lim _{\delta x \rightarrow 0} \frac{\delta x}{\delta t}}....(1)$
As $x$ and $y$ are differentiable function of $t$
$\lim _{\delta \rightarrow 00} \frac{\delta x}{\delta t}=\frac{d x}{d t} \text { and } \lim _{\delta t \rightarrow 0} \frac{\delta y}{\delta t}=\frac{d y}{d t}....(2)$
As $\text{R.H.S.}$ limit on $(1)$ exists and are finite $\text{L.H.S.}$ limit also exists and is finite $(2)$.
$\therefore \lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}....(3)$
But $\lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
Now, given
$x=\sin t, y=\cos t$
Differentiating w.r.t. $t$
$\therefore \frac{d x}{d t}=\cos t, \frac{d y}{d t}=-\sin t$
$\therefore \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{-\sin t}{\cos t}=-\tan t$

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Question 54 Marks

If $x=f(t)$ and $y=g(t)$ are differentiable function of $t$, then prove that $y$ is a differentlable function of $x$ and $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$, where $\frac{d x}{d t} \neq 0$. Hence find $\frac{d y}{d x}$ if $x=a \cos ^2 t$ and $y=a \sin ^2 t$

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Question 64 Marks

If $y=f(x)$ is a differentiable function of $x$ such that inverse function $x=f^{-1}(y)$ exists, then prove that $x$ is a differentlable function of $y$ and $\frac{d x}{d y}=\frac{1}{\frac{d y}{d x}}$ where $\frac{d y}{d x} \neq 0$. Hence find $\frac{d}{d x}\left(\tan ^{-1} x\right)$

Answer

Let $\delta y$ be the increment in $y$ corresponding to an increment $\delta x$ in $x$.
as $\delta x \rightarrow 0, \delta y \rightarrow 0$
Now y is a differentiable function of x.
$\therefore \lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}=\frac{d y}{d x}$
Now $\frac{\delta y}{\delta x} \times \frac{\delta x}{\delta y}=1$
$\therefore \frac{\delta x}{\delta y}=\frac{1}{\frac{\delta y}{\delta x}}$
Taking limits on both sides as$\delta x \rightarrow 0, w e \geq t$
$\lim _{\delta x \rightarrow 0} \frac{\delta x}{\delta y}=\lim _{\delta x \rightarrow 0}\left[\frac{1}{\frac{\delta y}{\delta x}}\right]=\frac{1}{\lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}}$
$\lim _{\delta x \rightarrow 0} \frac{\delta x}{\delta y}=\frac{1}{\lim _{d x \rightarrow 0} \frac{\delta y}{\delta x}} \quad \ldots .[$ as $\delta x \rightarrow 0, \delta y \rightarrow 0]$
Since limit in R.H.S. exists
limit in L.H.S. also exists and we have,
$\lim _{\delta y \rightarrow 0} \frac{\delta x}{\delta y}=\frac{d x}{d y}$
$\frac{d x}{d y}=\frac{1}{\frac{d y}{d x}}$, where $\frac{d y}{d x} \neq 0$
Let $y=\tan ^{-1} x$
$x=\tan y \Rightarrow \cos y=\frac{1}{\sqrt{1+\tan ^2 y}}=\frac{1}{\sqrt{1+x^2}}$
$\therefore \sec ^y \cdot \frac{d y}{d x}=1 \Rightarrow \frac{d x}{d y}=\sec ^2 y$
$\frac{d y}{d x}=\frac{1}{\frac{d x}{d y}}=\frac{1}{\sec ^2 y}=\cos ^2 y \Rightarrow \frac{d y}{d x}=\cos ^y$
$\frac{d\left(\tan ^{-1} x\right)}{d x}=\cos ^2 y=(\cos y)^2=\left(\frac{1}{\sqrt{1+x^2}}\right)^2$
$\therefore \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$

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Question 74 Marks

If $a x^2+2 h x y+b y^2=0$, show that $\frac{d^2 y}{d x^2}=0$

Answer

$a x^2+2 h x y+b y^2=0$.......(1)
Differentiate w.r.t. x
$2 a x+2 h x \frac{d y}{d x}+2 h y+2 b y \frac{d y}{d x}=0$
$\therefore a x+h x \frac{d y}{d x}+b y \frac{d y}{d x}+h y=0$
$(h x+b y) b \frac{y}{d x}=-1(a x+h y)$
$\therefore \frac{d y}{d x}=-\frac{a x+h y}{h x+b y}$.......(2)
From (1),we have
$\begin{aligned} & a x^2+h x y+h x y+b y^2=0 \\ & x(a x+h y)+y(h x+b y)=0 \\ & x(a x+h y)=-y(h x+b y)\end{aligned}$
$-\frac{a x+h y}{h x+b y}=\frac{y}{x}$
Put in (2),
$\frac{d y}{d x}=\frac{y}{x}$
Differentiate w.r.t. x
$\frac{d^2 y}{d x^2}=\frac{x \frac{d y}{d x}-y}{x^2}$
$=\frac{x\left(\frac{y}{x}\right)-y}{x^2}$
$=\frac{0}{x^2}$
$=0$

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Question 84 Marks

If $y=f(x)$ is a differentiable function of $x$, such that inverse function of $x=f^{-1}(y)$ exists, then prove that $x$ is a differentiable function of $y$ and $\frac{d x}{d y}=\frac{1}{\frac{d y}{d x}}$. where $\frac{d y}{d x} \neq 0$. Hence if $y=\sin ^{-1} x,-1 \leq x \leq 1, \frac{-\pi}{2} \leq y \leq \frac{\pi}{2}$ then show that $\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}}$ where $|x|<1$

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Maths Solve the Following Question.(4 Marks)

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