Higher Order Derivative

[eric_f]

Higher Order Derivative - Solved, thanks!

Hi all,

The homework problem states:

\(\displaystyle y(t)=20t^\frac{4}{5}-6t^\frac{2}{3}.\) Compute \(\displaystyle y'''(t).\)

I end up with \(\displaystyle y'''(t)=\frac{96}{25}t^\frac{-11}{5}-\frac{16}{9}t^\frac{-7}{3}.\)

The book's answer key gives \(\displaystyle y'''(t)=\frac{96}{25}t^\frac{-11}{15}-\frac{16}{9}t^\frac{-7}{3}.\)

The solutions manual gives \(\displaystyle y'''(t)=\frac{96}{25}t^\frac{-11}{25}-\frac{16}{9}t^\frac{-7}{3}.\)

...and that's where I'm at

Any idea on which one to put my faith in and rework the problem?

 

[DrPhil]

Hi all,

The homework problem states:

\(\displaystyle y(t)=20t^\frac{4}{5}-6t^\frac{2}{3}.\) Compute \(\displaystyle y'''(t).\)

I end up with \(\displaystyle y'''(t)=\frac{96}{25}t^\frac{-11}{5}-\frac{16}{9}t^\frac{-7}{3}.\)

The book's answer key gives \(\displaystyle y'''(t)=\frac{96}{25}t^\frac{-11}{15}-\frac{16}{9}t^\frac{-7}{3}.\)

The solutions manual gives \(\displaystyle y'''(t)=\frac{96}{25}t^\frac{-11}{25}-\frac{16}{9}t^\frac{-7}{3}.\)

...and that's where I'm at

Any idea on which one to put my faith in and rework the problem?

I agree with your answer:

\(\displaystyle y'''(t)=\frac{96}{25}t^\frac{-11}{5}-\frac{16}{9}t^\frac{-7}{3}.\)

I don't see how the denominator of the power in the first term changes when an integer is subtracted.

 

[eric_f]

I agree with your answer:

\(\displaystyle y'''(t)=\frac{96}{25}t^\frac{-11}{5}-\frac{16}{9}t^\frac{-7}{3}.\)

I don't see how the denominator of the power in the first term changes when an integer is subtracted.

That's what I was wondering. It just magically appears in the last step for some reason. Thanks for the help!

~Eric