Question

find the derivative of the function.
$f(t) = 8t^{2/3} - 7t^{1/3} + 1$
$f(t) = $
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Explanation:

Step1: Apply power rule to each term

Recall power rule: $\frac{d}{dt}(t^n)=nt^{n-1}$

Step2: Differentiate $8t^{2/3}$

$\frac{d}{dt}(8t^{2/3}) = 8\times\frac{2}{3}t^{\frac{2}{3}-1} = \frac{16}{3}t^{-1/3}$

Step3: Differentiate $-7t^{1/3}$

$\frac{d}{dt}(-7t^{1/3}) = -7\times\frac{1}{3}t^{\frac{1}{3}-1} = -\frac{7}{3}t^{-2/3}$

Step4: Differentiate constant term 1

$\frac{d}{dt}(1) = 0$

Step5: Combine all results

Add the derivatives of each term together.

Answer:

$\frac{16}{3}t^{-1/3} - \frac{7}{3}t^{-2/3}$ (or equivalently $\frac{16}{3t^{1/3}} - \frac{7}{3t^{2/3}}$)