How do I find the derivative of $f(x) = x tan^-1 - ln sqrt(1+x^2)$ ?

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Answer 1 · 2018-04-25T10:48:10

$tan^-1x$ .

$f(x)=xtan^-1x-lnsqrt(1+x^2)=xtan^-1x-ln(1+x^2)^(1/2)$ .

$:. f(x)=xtan^-1x-1/2ln(1+x^2)$ .

Using the usual rules of diffn., we get,

$f'(x)=x*d/dx{tan^-1x}+tan^-1x*d/dx{x}$

$-1/2*d/dx{ln(1+x^2)}$ ,

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

$=x/(1+x^2)+tan^-1x-1/cancel2*1/(1+x^2)*cancel(2)x$ ,

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

$rArr f'(x)=tan^-1x$ .

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How do I find the derivative of f(x) = x tan^-1 - ln sqrt(1+x^2) f(x) = x tan^-1 - ln sqrt(1+x^2)?