# 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

## 3 Answers

What you have is correct. You've applied the product rule correctly, and you've applied the chain rule correctly on the right term. Good job!

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,
$f'(x) = 8x^7\sqrt{5-3x} + x^8\left(-\frac{3}{2\sqrt{5-3x}}\right)$
Taking LCM of denominators,
$f'(x) = \dfrac{8x^7 2(5-3x) - 3x^8}{2\sqrt{5-3x}} = \dfrac{x^7(16(5-3x) - 3x) }{2\sqrt{5-3x}} = \dfrac{x^7(80-51x)}{2\sqrt{5-3x}}$