What is the derivative of \sqrt[3]{(t^3-3)} ?

The derivative of \sqrt[3]{(t^3-3)} is (t^2)/((t^3-3)^{2/3)}

What is the first derivative of \sqrt[3]{(t^3-3)} ?

The first derivative of \sqrt[3]{(t^3-3)} is (t^2)/((t^3-3)^{2/3)}

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derivative of

\frac{t ^{2}}{( t ^{3} - 3 ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( t ^{3} - 3 ) ^{\frac{2}{3}}} \frac{d}{d t} (t^{3} - 3)

= \frac{1}{3 ( t ^{3} - 3 ) ^{\frac{2}{3}}} \frac{d}{d t} (t^{3} - 3)

\frac{d}{d t} (t^{3} - 3) = 3 t^{2}

= \frac{1}{3 ( t ^{3} - 3 ) ^{\frac{2}{3}}} \cdot 3 t^{2}

Simplify

\frac{1}{3 ( t ^{3} - 3 ) ^{\frac{2}{3}}} \cdot 3 t^{2} : \frac{t ^{2}}{( t ^{3} - 3 ) ^{\frac{2}{3}}}

= \frac{t ^{2}}{( t ^{3} - 3 ) ^{\frac{2}{3}}}

What is the derivative of \sqrt[3]{(t^3-3)} ?
The derivative of \sqrt[3]{(t^3-3)} is (t^2)/((t^3-3)^{2/3)}
What is the first derivative of \sqrt[3]{(t^3-3)} ?
The first derivative of \sqrt[3]{(t^3-3)} is (t^2)/((t^3-3)^{2/3)}