How do you differentiate $\frac{x^{2} - 6 x + 9}{\sqrt{x - 3}}$ $(x^2 -6x + 9 )/ sqrt(x-3)$ using the quotient rule?

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How do you differentiate $\frac{x^{2} - 6 x + 9}{\sqrt{x - 3}}$ $(x^2 -6x + 9 )/ sqrt(x-3)$ using the quotient rule?

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Answer 1 · 2015-12-27T15:51:04

$f'(x) = ((2x-6)sqrt(x-3) - (x^2 - 6x + 9)(1/(2sqrt(x-3))))/(x-3)$

Explanation:

Let $f(x) = (x^2 - 6x + 9)/sqrt(x-3)$ .

The quotient rule tells us that the derivative of $(u(x))/(v(x))$ is $(u'(x)v(x) - u(x)v'(x))/(v(x)^2)$ . Here, let $u(x) = x^2 - 6x + 9$ and $v(x) = sqrt(x-3)$ . So $u'(x) = 2x - 6$ and $v'(x) = 1/(2sqrt(x-3))$ .

We now apply the quotient rule.

$f'(x) = ((2x-6)sqrt(x-3) - (x^2 - 6x + 9)(1/(2sqrt(x-3))))/(x-3)$

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How do you differentiate $\frac{x^{2} - 6 x + 9}{\sqrt{x - 3}}$ $(x^2 -6x + 9 )/ sqrt(x-3)$ using the quotient rule?

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How do you differentiate $\frac{x^{2} - 6 x + 9}{\sqrt{x - 3}}$ $(x^2 -6x + 9 )/ sqrt(x-3)$ using the quotient rule?