Question

D) Differentiate each of the following funcion
1+√x/1-√x
​

Answer 1

Answer:

here's your answer to the question

Answer 2

Given,

$\displaystyle\sf{\longrightarrow u=\dfrac {1+\sqrt x}{1-\sqrt x}}$

Substitute,

$\displaystyle\sf{\longrightarrow \sqrt x=\cos (2\theta)}$

$\displaystyle\sf{\longrightarrow x=\cos^2(2\theta)}$

$\displaystyle\sf{\longrightarrow \dfrac {dx}{d\theta}=-4\sin(2\theta)\cos (2\theta)}$

Therefore,

$\displaystyle\sf{\longrightarrow u=\dfrac {1+\cos (2\theta)}{1-\cos(2\theta)}}$

$\displaystyle\sf{\longrightarrow u=\dfrac {2\cos^2\theta}{2\sin^2\theta}}$

$\displaystyle\sf{\longrightarrow u=\dfrac {\cos^2\theta}{\sin^2\theta}}$

By Quotient Rule,

$\displaystyle\sf{\longrightarrow \dfrac {du}{d\theta}=\dfrac {-\sin (2\theta)}{\sin^4\theta}}$

Hence, by Chain Rule,

$\displaystyle\sf{\longrightarrow\dfrac {du}{dx}=\dfrac {du}{d\theta}\cdot\dfrac {d\theta}{dx}}$

$\displaystyle\sf{\longrightarrow\dfrac {du}{dx}=\dfrac {-\sin (2\theta)}{\sin^4\theta}\cdot\dfrac {1}{-4\sin (2\theta)\cos (2\theta)}}$

$\displaystyle\sf{\longrightarrow\dfrac {du}{dx}=\dfrac {1}{4\sin^4\theta}\cdot\dfrac {1}{\cos (2\theta)}}$

$\displaystyle\sf{\longrightarrow\dfrac {du}{dx}=\dfrac {1}{\cos (2\theta)(1-\cos(2\theta))^2}}$

Undoing $\displaystyle\sf {\sqrt x=\cos (2\theta),}$

$\displaystyle\underline {\underline {\sf{\longrightarrow\dfrac {du}{dx}=\dfrac {1}{\sqrt x(1-\sqrt x)^2}}}}$