Trouble taking the derivative. Hung up on the algebra

I don't know if I should post this here or on the algebra forum.

Trying to find the derivative of f (X) = ³√t (t² + 4)

Let ³√t be a

Let (t² + 4) be b

f(X)' = a'b +b'a

a' =⅓t٨-⅔

b' = 2t

thus f(x)' =(⅓t٨-⅔)(t² + 4) + 2t ³√t

or (t²+4/3t٨⅔) +2t*t٨⅓

I do not understand how to get from here to the answer (7t+4)/(3t٨⅔)

Thanks for any help or advice.

Re: Trouble taking the derivative. Hung up on the algebra

$\displaystyle t^{\frac{1}{3}}(t^2 + 4)$

distribute ...

$\displaystyle t^{\frac{7}{3}} + 4t^{\frac{1}{3}}$

take the derivative w/r to t using the power rule ...

$\displaystyle \frac{7}{3} t^{\frac{4}{3}} + \frac{4}{3} t^{-\frac{2}{3}}$

$\displaystyle \frac{7t^{\frac{4}{3}}}{3} + \frac{4}{3t^{\frac{2}{3}}}$

common denominator is $\displaystyle 3t^{\frac{2}{3}}$ ...

$\displaystyle \frac{7t^{\frac{4}{3}} \cdot t^{\frac{2}{3}}}{3t^{\frac{2}{3}}} + \frac{4}{3t^{\frac{2}{3}}}$

$\displaystyle \frac{7t^2+4}{3t^{\frac{2}{3}}}$

Re: Trouble taking the derivative. Hung up on the algebra

Thank you for the help. Evidently, the fundamental problem is that I do not understand dealing with these fractional powers. I have searched it on the web and gained some knowledge, but why t⅓ X t² is t٨7/3 alludes me. Can you explain or direct me to an explanation? Also, where do you find the fonts to write out this math so elegantly? (At 71, I am finding it a struggle to re-tool in math, so I can go on an learn some calculus!)

Re: Trouble taking the derivative. Hung up on the algebra

you should already know the fundamental rule that $\displaystyle t^a \cdot t^b = t^{a+b}$

$\displaystyle t^{\frac{1}{3}} \cdot t^2 = t^{\frac{1}{3}} \cdot t^{\frac{6}{3}} = t^{\frac{1}{3} + \frac{6}{3}} = t^{\frac{7}{3}}$