Question

Derivative of tan^-1(1+2x)/(1-2x) wrt ✓1+4x​

Answer 1

Let,

$\longrightarrow u=\tan^{-1}\left(\dfrac{1+2x}{1-2x}\right)\quad\quad\dots(1)$

Substitute,

$\longrightarrow v=\sqrt{1+4x}$

Thus the derivative of $\tan^{-1}\left(\dfrac{1+2x}{1-2x}\right)$ with respect to $\sqrt{1+4x}$ will be $\dfrac{du}{dv}.$

$\longrightarrow v^2=1+4x$

$\longrightarrow x=\dfrac{v^2-1}{4}$

Then (1) becomes,

$\longrightarrow u=\tan^{-1}\left(\dfrac{1+2\cdot\frac{v^2-1}{4}}{1-2\cdot\frac{v^2-1}{4}}\right)$

$\longrightarrow u=\tan^{-1}\left(\dfrac{1+\frac{v^2-1}{2}}{1-\frac{v^2-1}{2}}\right)$

$\longrightarrow u=\tan^{-1}\left(\dfrac{2+v^2-1}{2-v^2+1}\right)$

$\longrightarrow u=\tan^{-1}\left(\dfrac{v^2+1}{3-v^2}\right)\quad\quad\dots(2)$

Let,

$\longrightarrow w=\dfrac{v^2+1}{3-v^2}\quad\quad\dots(3)$

Differentiating wrt $v,$

$\longrightarrow \dfrac{dw}{dv}=\dfrac{d}{dv}\left(\dfrac{v^2+1}{3-v^2}\right)$

By quotient rule,

$\longrightarrow \dfrac{dw}{dv}=\dfrac{2v(3-v^2)+2v(v^2+1)}{(3-v^2)^2}$

$\longrightarrow \dfrac{dw}{dv}=\dfrac{2v(3-v^2+v^2+1)}{(3-v^2)^2}$

$\longrightarrow \dfrac{dw}{dv}=\dfrac{8v}{(3-v^2)^2}\quad\quad\dots(4)$

Then (2) becomes,

$\longrightarrow u=\tan^{-1}\left(w\right)$

Differentiating wrt $v,$

$\longrightarrow \dfrac{du}{dv}=\dfrac{d}{dv}\big(\tan^{-1}\left(w\right)\big)$

By chain rule,

$\longrightarrow \dfrac{du}{dv}=\dfrac{d}{dw}\big(\tan^{-1}\left(w\right)\big)\cdot\dfrac{dw}{dv}$

$\longrightarrow \dfrac{du}{dv}=\dfrac{1}{1+w^2}\cdot\dfrac{dw}{dv}$

From (3) and (4),

$\longrightarrow \dfrac{du}{dv}=\dfrac{1}{1+\left(\dfrac{v^2+1}{3-v^2}\right)^2}\cdot\dfrac{8v}{(3-v^2)^2}$

$\longrightarrow \dfrac{du}{dv}=\dfrac{(3-v^2)^2}{(v^2+1)^2+(3-v^2)^2}\cdot\dfrac{8v}{(3-v^2)^2}$

$\longrightarrow \dfrac{du}{dv}=\dfrac{8v}{(v^2+1)^2+(3-v^2)^2}$

Undoing substitution $v=\sqrt{1+4x}\,,$

$\longrightarrow \dfrac{du}{dv}=\dfrac{8\sqrt{1+4x}}{(1+4x+1)^2+(3-1-4x)^2}$

$\longrightarrow \dfrac{du}{dv}=\dfrac{8\sqrt{1+4x}}{(2+4x)^2+(2-4x)^2}$

Since $(a+b)^2+(a-b)^2=2(a^2+b^2),$

$\longrightarrow \dfrac{du}{dv}=\dfrac{8\sqrt{1+4x}}{2(4+16x^2)}$

$\longrightarrow\underline{\underline{\dfrac{du}{dv}=\dfrac{\sqrt{1+4x}}{1+4x^2}}}$

This is the answer to the question.