Question

Differentiate log (1 + x) w.r.t. sin-1x.

Answer 1

Let,

$\small\longrightarrow u=\log(1+x)$

$\small\longrightarrow\dfrac{du}{dx}=\dfrac{d}{dx}[\log(1+x)]$

$\small\text{ \longrightarrow\dfrac{du}{dx}=\dfrac{\dfrac{d}{dx}[1+x]}{1+x} }$

$\small\longrightarrow\dfrac{du}{dx}=\dfrac{1}{1+x}\quad\dots(1)$

and,

$\small\longrightarrow v=\sin^{-1}x$

$\small\longrightarrow\dfrac{dv}{dx}=\dfrac{d}{dx}[\sin^{-1}x]$

$\small\text{ \longrightarrow\dfrac{dv}{dx}=\dfrac{1}{\sqrt{1-x^2}}\quad\dots(2) }$

We need to find $\small\dfrac{du}{dv}.$

Dividing numerator and denominator of $\small\dfrac{du}{dv}$ by dx we get,

$\small\text{ \longrightarrow\dfrac{du}{dv}=\dfrac{\left(\dfrac{du}{dx}\right)}{\left(\dfrac{dv}{dx}\right)} }$

From (1) and (2),

$\small\text{ \longrightarrow\dfrac{du}{dv}=\dfrac{\left(\dfrac{1}{1+x}\right)}{\left(\dfrac{1}{\sqrt{1-x^2}}\right)} }$

$\small\text{ \longrightarrow\dfrac{du}{dv}=\dfrac{\sqrt{1-x^2}}{1+x} }$

$\small\text{ \longrightarrow\dfrac{du}{dv}=\dfrac{\sqrt{(1-x)(1+x)}}{1+x} }$

$\small\text{ \longrightarrow\dfrac{du}{dv}=\dfrac{\sqrt{1-x}\cdot\sqrt{1+x}}{1+x} }$

$\small\text{ \longrightarrow\dfrac{du}{dv}=\dfrac{\sqrt{1-x}}{\sqrt{1+x}} }$

$\small\text{ \longrightarrow\underline{\underline{\dfrac{du}{dv}=\sqrt{\dfrac{1-x}{1+x}}}} }$

Answer 2

Answer:

The derivative is √[(1-x)/(1+x)]

HOPE IT HELPS!!

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