# Derivative: Square Root

I wish to find the derivative of the following function:

$$f(x)=x^8\sqrt{5-3x}$$

So far I've used the product rule to come up with the following...

$$8x^7\sqrt{5-3x} + x^8\left(-\frac{3}{2\sqrt{5-3x}}\right)$$

But from there I'm completely stuck. Would I change the sqrt to the exponent -1/2 and then use the chain rule? Thanks in advance, step by step instructions would be awesome.

• – lab bhattacharjee Feb 20 '14 at 18:50
• Why do you think there's something wrong with what you already have? – Thomas Russell Feb 20 '14 at 18:51
• The assignment has multiple choice answers and that isn't one of them – user102817 Feb 20 '14 at 18:53
• try rationalizing the denominator, putting everything over the same denomenator, and/or factoring – David Diaz Jun 14 '18 at 13:37

If your choices are missing square roots, then do as Ross suggested and multiply top and bottom by $\sqrt{5-3x}$.
You are done. When you put the square root in the denomimator you changed the exponent to $\frac {-1}2$ You could "simplify" the second term by multiplying numerator and denominator by $\sqrt {5-3x}$ to get the square root out of the denominator. Some would find that a preferable expression, some would not.
$f(x) = x^8 \sqrt{5-3x}$
Differentiating w.r.t. $x$ and as you have done by using chain rule,
$f'(x) = 8x^7\sqrt{5-3x} + x^8\left(-\frac{3}{2\sqrt{5-3x}}\right)$
$f'(x) = \dfrac{8x^7 2(5-3x) - 3x^8}{2\sqrt{5-3x}} = \dfrac{x^7(16(5-3x) - 3x) }{2\sqrt{5-3x}} = \dfrac{x^7(80-51x)}{2\sqrt{5-3x}}$