What is the derivative of $y = {ln tan}^{- 1} x$ $y= Ln tan^-1 x$ ?

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Answer 1 · 2018-05-28T14:24:47

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

Given: $y = ln ({tan}^{- 1} x)$ $y=ln(tan^-1x)$ .

Use the chain rule, which states that,

$\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $dy/dx=dy/(du)*(du)/dx$

Let $u=tan^-1x,:.(du)/dx=1/(x^2+1)$ .

Then, $y=lnu,:.dy/(du)=1/u$ .

Combining, we get:

$dy/dx=1/u*1/(x^2+1)$

$=1/(u(x^2+1))$

Replacing back $u=tan^-1x$ , we get:

$=1/((tan^-1x)(x^2+1))$

What is the derivative of y= Ln tan^-1 xy=lntan−1xy= Ln tan^-1 x?