1. Sep 24, 2009

fghtffyrdmns

1. The problem statement, all variables and given/known data
Differentiate using the product rule

2. Relevant equations

$$(2t^2+t^{(1/3)})(4t-5)$$

3. The attempt at a solution

$$h'(t) = f'(t)g(t)+f(t)g'(t)$$

$$(4t+\frac {1}{3}t^{\frac {-2}{3}})(4t-5)+(2t^2+t^{(1/3)})(4)$$
$$\frac {-5}{3t^{2/3}}+24t^{2}-20t+\frac {16}{3}t^{1/3}$$

Why is this wrong?

Last edited: Sep 24, 2009
2. Sep 24, 2009

anubis01

maybe your missing something in the question because the work you showed is correct.

3. Sep 24, 2009

Pengwuino

Looks right as can be.

4. Sep 24, 2009

fghtffyrdmns

The answer in the book is much different, though:
$$h'(t) = \frac {72x^{8/3} - 60x^{5/3} + 16x -5}{3x^{2/3}}$$

Something is not right :/.

5. Sep 24, 2009

Pengwuino

Those 2 answers are equal....... other then the t's all of a sudden being x's :).

6. Sep 24, 2009

Dick

It's not that much different. 72*x^(8/3)/(3*x^(2/3)) is 24*x^2 which if I replace x by t corresponds to the 24*t^2 in your solution. Can you match the other terms up as well? They just factored the answer in a different way.

7. Sep 24, 2009

fghtffyrdmns

Ahhhhh, that explains it! Thank you very much :).