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)#