Find the derivative of the function y=x² /arctgx ?

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Find the derivative of the function y=x² /arctgx ?

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Answer 1 · 2018-06-14T20:43:07

$\frac{d y}{d x} = {(arctan x)}^{- 2} (2 x arctan x - \frac{x^{2}}{x^{2} + 1})$ $dy/dx=(arctanx)^-2(2xarctanx-x^2/(x^2+1))$

Explanation:

$y = \frac{x^{2}}{arctan x}$ $y=x^2/arctanx$

$y = x^{2} {(arctan x)}^{- 1}$ $y=x^2(arctanx)^-1$

Let's differentiate this using the product rule.

$\frac{d y}{d x} = (\frac{d}{d x} x^{2}) {(arctan x)}^{- 1} + x^{2} (\frac{d}{d x} {(arctan x)}^{- 1})$ $dy/dx=(d/dxx^2)(arctanx)^-1+x^2(d/dx(arctanx)^-1)$

The derivative of $x^{2}$ $x^2$ can be found with the product rule. To find the derivative of ${(arctan x)}^{- 1}$ $(arctanx)^-1$ , use the chain rule.

$\frac{d y}{d x} = 2 x {(arctan x)}^{- 1} + x^{2} (- {(arctan x)}^{- 2}) (\frac{d}{d x} arctan x)$ $dy/dx=2x(arctanx)^-1+x^2(-(arctanx)^-2)(d/dxarctanx)$

$\frac{d y}{d x} = 2 x {(arctan x)}^{- 1} - x^{2} {(arctan x)}^{- 2} (\frac{1}{x^{2} + 1})$ $dy/dx=2x(arctanx)^-1-x^2(arctanx)^-2(1/(x^2+1))$

$\frac{d y}{d x} = {(arctan x)}^{- 2} (2 x arctan x - \frac{x^{2}}{x^{2} + 1})$ $dy/dx=(arctanx)^-2(2xarctanx-x^2/(x^2+1))$

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Find the derivative of the function y=x² /arctgx ?

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