What is the derivative of $(tan^-1 (x+2)/(1+2x))$ ?

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Answer 1 · 2018-07-18T17:04:13

$dy/dx=frac{1}{(1+2x)(x^2+4x+5)}-2/{(1+2x)^2} \tan^{-1}(x+2)$

Given function:

$y=\frac{\tan^{-1}(x+2)}{1+2x}$

differentiating w.r.t. $x$ using product rule as follows

$dy/dx=\frac{d}{dx}(\frac{\tan^{-1}(x+2)}{1+2x})$

$=\frac{(1+2x)d/dx\tan^{-1}(x+2)-\tan^{-1}(x+2)d/dx(1+2x)}{(1+2x)^2}$

$=\frac{(1+2x)\frac{1}{1+(x+2)^2}-\tan^{-1}(x+2)(2)}{(1+2x)^2}$

$=frac{1}{(1+2x)(x^2+4x+5)}-2/{(1+2x)^2} \tan^{-1}(x+2)$

What is the derivative of (tan^-1 (x+2)/(1+2x))(tan^-1 (x+2)/(1+2x))?