How do you find the derivative of $x^{2} \cdot {tan}^{- 1} x$ $x^2*tan^-1 x$ ?

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How do you find the derivative of $x^{2} \cdot {tan}^{- 1} x$ $x^2*tan^-1 x$ ?

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Answer 1 · 2016-05-09T09:29:46

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

Explanation:

Differentiate using the $product rule$ $color(blue)" product rule "$

• If f(x) = g(x).h(x) then f'(x) = g(x).h'(x) + h(x).g'(x)
$---------------------------------------------------------$ $"---------------------------------------------------------"$
here $g(x)=x^2rArrg'(x)=2x$

and $h(x)=tan^-1xrArrh'(x)=1/(1+x^2)$
$"---------------------------------------------------------"$
Substitute these values into f'(x)

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

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

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How do you find the derivative of $x^{2} \cdot {tan}^{- 1} x$ $x^2*tan^-1 x$ ?

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