If `y = (ln(tan^-1 x))^2` , find `dy/dx` ?

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The function `y = (ln(tan^-1 x))^2` .

For `f(x) = g(h(x))` , `f'(x) = g'(h(x))*h'(x)` .

For the given function:

`dy/dx = 2*ln(tan^-1 x)*(1/(tan^-1 x))*1/(x^2+1) `

= `(2*ln(tan^-1 x))/(tan^-1 x*(x^2+1))`

The required derivative of `y = (ln(tan^-1 x))^2` is `dy/dx = (2*ln(tan^-1 x))/(tan^-1 x*(x^2+1))`

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`y=[log(tan^-1 x)]^2`

`dy/dx = 2 log (tan^-1 x) d/dx (log (tan^-1 x))`

       `=2 log (tan^-1 x) (1/(tan^-1 (x))) d/dx (tan^-1 x)`

       `= 2 log (tan^-1 x) (1/(tan^-1 (x))) (1/(1+x^2))`

     ` =(2 log (tan^-1 (x))) / ((1+x^2) tan^-1 (x))`

So the Answer is ` (2 log (tan^-1 (x))) / ((1+x^2) tan^-1 (x))`

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