What is the derivative of $y = \frac{5 x}{\sqrt{x^{2} + 9}}$ $y= (5x)/sqrt (x^2+9)$ ?

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Answer 1 · 2017-04-13T15:43:24

$y'=45/(x^2+9)^(3/2)$

By rewriting a bit,

$y=5 x/(x^2+9)^(1/2)$

$y'=5(1 cdot (x^2+9)^(1/2)-x cdot1/(cancel2)(x^2+9)^(-1/2)(cancel 2x))/((x^2+9)^(1/2))^2$

By cleaning up a bit,

$=5((x^2+9)^(1/2)-x^2/(x^2+9)^(1/2))/(x^2+9)$

By multiply the numerator and the denominator by $(x^2+9)^(1/2)$ ,

$=5(cancel(x^2)+9-cancel(x^2))/(x^2+9)^(3/2)=45/(x^2+9)^(3/2)$

I hope that this was clear.

What is the derivative of y= (5x)/sqrt (x^2+9)y=5x√x2+9y= (5x)/sqrt (x^2+9)?