Product rule differentiation of (2t² + t^(1/3))(4t - 5)

[fghtffyrdmns]

Homework Statement

Differentiate using the product rule

Homework Equations

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

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?

 

[anubis01]

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

 

[Pengwuino]

Looks right as can be.

 

[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 :/.

 

[Pengwuino]

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

 

[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.

 

[fghtffyrdmns]

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