Find the derivative of $-8t/\sqrt[3]{t^2}$
How do you get the answer $\dfrac{-8t^{-2}}3$ ?
Find the derivative of $-8t/\sqrt[3]{ t^2}$
0
$\begingroup$
calculus
derivatives
-
0Hint: $\frac{t}{\sqrt[3]{t^2}}=t^{1/3}$ – 2017-02-04
-
0Did you forget the denominator in the exponent? The answer should be $-\frac{8}{3}t^{-\frac{2}{3}}$. – 2017-02-04
-
0The title is different from the body of question. – 2017-02-04
-
0Anna- Sorry thats what I meant to post as the answer – 2017-02-04
1 Answers
3
Assuming there is no typo:
We have $\sqrt [3]{t^2} = t^{\frac {2}{3}} $. Thus we get $$f (t) = \frac {-8t}{t^{\frac {2}{3}}} = -8t^{1-\frac {2}{3}} = -8t^{\frac {1}{3}} $$ Thus the derivative comes out to be $$\boxed {-8\frac {1}{3}t^{\frac {1}{3}-1} = -\frac {8}{3} t^{-\frac {2}{3}}} $$ Hope it helps.
-
0Thanks I don't know why I thought I was done when I got to -8t^1/3 – 2017-02-04