Question

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

Answer 1

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

Explanation:

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)`