Question

dy/dx of tan^-1(((a/b)-tanx)/(1+a/btanx))

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:y = {tan}^{ - 1} \bigg(\dfrac{\dfrac{a}{b} \: - \: tanx}{1 + \dfrac{a}{b} \: tanx } \bigg)$

We know,

$\boxed{ \bf{ \: {tan}^{ - 1}\bigg(\dfrac{x - y}{1 + xy}\bigg) = {tan}^{ - 1}x - {tan}^{ - 1}y}}$

So, using this identity, we get

$\rm :\longmapsto\:y = {tan}^{ - 1}\bigg(\dfrac{a}{b} \bigg) - {tan}^{ - 1}(tanx)$

$\rm :\longmapsto\:y = {tan}^{ - 1}\bigg(\dfrac{a}{b} \bigg) - x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}\bigg( {tan}^{ - 1}\bigg(\dfrac{a}{b} \bigg) - x\bigg)$

We know,

$\boxed{ \bf{ \:\dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u \: + \: \dfrac{d}{dx}v}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{d}{dx}{tan}^{ - 1}\bigg(\dfrac{a}{b} \bigg) - \dfrac{d}{dx}x$

We know,

$\boxed{ \bf{ \:\dfrac{d}{dx}k = 0}}$

and

$\boxed{ \bf{ \:\dfrac{d}{dx}x = 1}}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{dy}{dx}= 0 - 1$

$\rm :\longmapsto\:\boxed{ \: \: \: \: \: \bf{ \:\dfrac{dy}{dx} \: = \: - \: 1 \: \: \: \: \: }}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cox & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$