y=x^2 1+x dydx= ddxx^21+x ( uv )^'=vu^' uv^'v^2 = (1+x)(2x) x^2(1)(1+x)^2 = 2x^2+2x x^2(1+x)^2 = x^2 +2x(1+x)^2 ...

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Algebra of Derivatives

y=x21+x Find ...

Question

$y = \frac{x^{2}}{1 + x}$ Find $\frac{d y}{d x}$

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Solution

$y = \frac{x^{2}}{1 + x}$

$\frac{d y}{d x} = \frac{d}{d x} \frac{x^{2}}{1 + x}$

${(\frac{u}{v})}^{'} = \frac{v u^{'} - u v^{'}}{v^{2}}$

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

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

$= \frac{x^{2} + 2 x}{(1 + x)^{2}}$

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