Simplifying Differential for f(x) = x^(4/5) (x - 4)^2

[mattcl]

Hey guys, first time poster so please go easy on me!

The original question is, 'find the critical numbers of the function' f(x) = x^(4/5) (x-4)^2

Ive gotten the f'(x) to the point 2(x-4)x^(4/5) + (4x-4)^2 / 5 root5(x)


but Im having trouble simplifying to

2(x-4)(7x-8) / 5root5(x)​

Ive tried putting 5 rootx onto the left side denominator and numerator and adding but Im getting confused simplifying down. hoping someone could clarify for me.

Thanks.

 

[ksdhart]

Hey guys, first time poster so please go easy on me!

The original question is, 'find the critical numbers of the function' f(x) = x^(4/5) (x-4)^2

Ive gotten the f'(x) to the point 2(x-4)x^(4/5) + (4x-4)^2 / 5 root5(x)


but Im having trouble simplifying to

2(x-4)(7x-8) / 5root5(x)​

Ive tried putting 5 rootx onto the left side denominator and numerator and adding but Im getting confused simplifying down. hoping someone could clarify for me.

Thanks.

Well, the first thing I'd note is that your initial expression for the derivative is wrong. I think it's just a typo though. What you wrote is:

$\displaystyle \displaystyle 2\left(x-4\right)x^{\frac{4}{5}}+\frac{\left(4x-4\right)^2}{5\sqrt[5]{x}}$

But in the actual derivative, the 4 should be on the outside of the parentheses, like so:

$\displaystyle \displaystyle 2\left(x-4\right)x^{\frac{4}{5}}+\frac{4\left(x-4\right)^2}{5\sqrt[5]{x}}$

With that minor correction out of the way, using the method you said you tried, a bit of simplification yields:

$\displaystyle \displaystyle \left(x-4\right)\left(\frac{2x^{\frac{4}{5}}\cdot 5\sqrt[5]{x}}{5\sqrt[5]{x}}+\frac{4\left(x-4\right)}{5\sqrt[5]{x}}\right)$

Try continuing from here and see what you get. As a hint, remember that the fifth root of x can also be written as x^(1/5).

 

[mattcl]

yes i got it! simple simple simple! Thanks alot for your time.