Question

answerthis by step by step explanation​

Answer 1

Question:

Differentiate:

$\displaystyle{\sf\:\tan^{-\:1}\:\left(\:\dfrac{5x\:+\:1}{3\:-\:x\:-\:6x^2}\:\right)}$

Answer:

$\displaystyle{\boxed{\red{\sf\:\dfrac{d}{dx}\:\left[\:\tan^{-\:1}\:\left(\:\dfrac{5x\:+\:1}{3\:-\:x\:-\:6x^2}\:\right)\:\right]\:=\:\dfrac{3}{9x^2\:+\:12x\:+\:5}\:+\:\dfrac{1}{2x^2\:-\:2x\:+\:1}\:}}}$

Step-by-step-explanation:

We have given an inverse trigonometric function.

We have to find the derivative of that function.

The given inverse trigonometric function is

$\displaystyle{\sf\:\tan^{-\:1}\:\left(\:\dfrac{5x\:+\:1}{3\:-\:x\:-\:6x^2}\:\right)}$

Let this function be y.

$\displaystyle{\therefore\:\sf\:y\:=\:\tan^{-\:1}\:\left(\:\dfrac{5x\:+\:1}{3\:-\:x\:-\:6x^2}\:\right)}$

$\displaystyle{\implies\sf\:y\:=\:\tan^{-\:2}\:\left(\:\dfrac{5x\:+\:1}{1\:+\:2\:-\:x\:-\:6x^2}\:\right)}$

$\displaystyle{\implies\sf\:y\:=\:\tan^{-\:1}\:\left(\:\dfrac{5x\:+\:1}{1\:+\:(\:2\:-\:4x\:+\:3x\:-\:6x^2\:)}\:\right)}$

$\displaystyle{\implies\sf\:y\:=\:\tan^{-\:1}\:\left(\:\dfrac{5x\:+\:1}{1\:+\:[\:2\:(\:1\:-\:2x\:)\:+\:3x\:(\:1\:-\:2x\:)\:]}\:\right)}$

$\displaystyle{\implies\sf\:y\:=\:\tan^{-\:1}\:\left(\:\dfrac{5x\:+\:1}{1\:+\:[\:(\:1\:-\:2x\:)\:(\:2\:+\:3x\:)\:]}\:\right)}$

$\displaystyle{\implies\sf\:y\:=\:\tan^{-\:1}\:\left(\:\dfrac{5x\:+\:1}{1\:+\:(\:3x\:+\:2\:)\:(\:-\:2x\:+\:1\:)}\:\right)}$

$\displaystyle{\implies\sf\:y\:=\:\tan^{-\:1}\:\left(\:\dfrac{3x\:+\:2x\:+\:2\:-\:1}{1\:+\:(\:3x\:+\:2\:)\:(\:-\:2x\:+\:1\:)}\:\right)}$

$\displaystyle{\implies\sf\:y\:=\:\tan^{-\:1}\:\left(\:\dfrac{(\:3x\:+\:2\:)\:+\:(\:2x\:-\:1\:)}{1\:+\:(\:3x\:+\:2\:)\:(\:-\:2x\:+\:1\:)}\:\right)}$

$\displaystyle{\implies\sf\:y\:=\:\tan^{-\:1}\:\left(\:\dfrac{(\:3x\:+\:2\:)\:-\:(\:-\:2x\:+\:1\:)}{1\:+\:(\:3x\:+\:2\:)\:(\:-\:2x\:+\:1\:)}\:\right)}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:\tan^{-\:1}\:a\:-\:\tan^{-\:1}\:b\:=\:\tan^{-\:1}\:\left(\:\dfrac{a\:-\:b}{1\:+\:ab}\:\right)\:}}}$

$\displaystyle{\implies\sf\:y\:=\:\tan^{-\:1}\:(\:3x\:+\:2\:)\:-\:\tan^{-\:1}\:(\:-\:2x\:+\:1\:)}$

Differentiating both sides w.r.t. x, we get,

$\displaystyle{\sf\:\dfrac{d}{dx}\:(\:y\:)\:=\:\dfrac{d}{dx}\:\left[\:\tan^{-\:1}\:(\:3x\:+\:2\:)\:-\:\tan^{-\:1}\:(\:-\:2x\:+\:1\:)\:\right]}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:\dfrac{d}{dx}\:(\:u\:-\:v\:)\:=\:\dfrac{du}{dx}\:-\:\dfrac{dv}{dx}\:}}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{d}{dx}\:\left[\:\tan^{-\:1}\:(\:3x\:+\:2\:)\:\right]\:-\:\dfrac{d}{dx}\:\left[\:\tan^{-\:1}\:(\:-\:2x\:+\:1\:)\:\right]}$

We know that,

$\displaystyle{\boxed{\green{\sf\:\dfrac{d}{dx}\:[\:\tan^{-\:1}\:(\:u\:)\:]\:=\:\dfrac{1}{1\:+\:u^2}\:}}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{1}{1\:+\:(\:3x\:+\:2\:)^2}\:.\:\dfrac{d}{dx}\:\left(\:3x\:+\:2\:\right)\:-\:\dfrac{1}{1\:+\:(\:-\:2x\:+\:1\:)^2}\:.\:\dfrac{d}{dx}\:(\:-\:2x\:+\:1\:)}$

$\displaystyle{\:\implies}\sf\:\dfrac{dy}{dx}\:=\:\dfrac{1}{1\:+\:(\:3x\:+\:2\:)^2}\:.\:\left[\:\dfrac{d}{dx}\:(\:3x\:)\:+\:\dfrac{d}{dx}\:(\:2\:)\:\right]\:-\:\dfrac{1}{1\:+\:(\:-\:2x\:+\:1\:)^2}\:.\:\left[\:\dfrac{d}{dx}\:(\:-\:2x\:)\:+\:\dfrac{d}{dx}\:(\:1\:)\:\right]$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{1}{1\:+\:(\:3x\:+\:2\:)^2}\:.\:(\:3\:+\:0\:)\:-\:\dfrac{1}{1\:+\:(\:-\:2x\:+\:1\:)^2}\:.\:(\:-\:2\:+\:0\:)}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{3}{1\:+\:(\:3x\:+\:2\:)^2}\:-\:\dfrac{-\:2}{1\:+\:(\:-\:2x\:+\:1\:)^2}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{3}{1\:+\:(\:3x\:)^2\:+\:2\:\times\:3x\:\times\:2\:+\:2^2}\:-\:\dfrac{-\:2}{1\:+\:(\:-\:2x\:)^2\:+\:2\:\times\:(\:-\:2x\:)\:\times\:1\:+\:1^2}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{3}{1\:+\:9x^2\:+\:12x\:+\:4}\:-\:\dfrac{-\:2}{1\:+\:4x^2\:-\:4x\:+\:1}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{3}{9x^2\:+\:12x\:+\:4\:+\:1}\:+\:\dfrac{2}{4x^2\:-\:4x\:+\:1\:+\:1}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{3}{9x^2\:+\:12x\:+\:5}\:+\:\dfrac{2}{4x^2\:-\:4x\:+\:2}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{3}{9x^2\:+\:12x\:+\:5}\:+\:\dfrac{\cancel{2}}{\cancel{2}\:(\:2x^2\:-\:2x\:+\:1\:)}}$

$\displaystyle{\implies\sf\:\dfrac{dy}{dx}\:=\:\dfrac{3}{9x^2\:+\:12x\:+\:5}\:+\:\dfrac{1}{2x^2\:-\:2x\:+\:1}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\dfrac{d}{dx}\:\left[\:\tan^{-\:1}\:\left(\:\dfrac{5x\:+\:1}{3\:-\:x\:-\:6x^2}\:\right)\:\right]\:=\:\dfrac{3}{9x^2\:+\:12x\:+\:5}\:+\:\dfrac{1}{2x^2\:-\:2x\:+\:1}\:}}}}$