What is the derivative of \sqrt[3]{t^2}-4sqrt(t^3) ?

The derivative of \sqrt[3]{t^2}-4sqrt(t^3) is (2t)/(3(t^2)^{2/3)}-6sqrt(t)

What is the first derivative of \sqrt[3]{t^2}-4sqrt(t^3) ?

The first derivative of \sqrt[3]{t^2}-4sqrt(t^3) is (2t)/(3(t^2)^{2/3)}-6sqrt(t)

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derivative of

\frac{2 t}{3 ( t ^{2} ) ^{\frac{2}{3}}} - 6 t

Solution steps

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

\frac{d}{d t} (4 t^{3}) = 6 t

= \frac{2 t}{3 ( t ^{2} ) ^{\frac{2}{3}}} - 6 t

x

\frac{d}{d x} (x ln (x^{18}))

\int \frac{x ^{2} + 3 x - 1}{x + 2} d x

\int_{0}^{3} (8 - 2 x) d x

\frac{\partial}{\partial y} (sin (x y) + x sin (y))

What is the derivative of \sqrt[3]{t^2}-4sqrt(t^3) ?
The derivative of \sqrt[3]{t^2}-4sqrt(t^3) is (2t)/(3(t^2)^{2/3)}-6sqrt(t)
What is the first derivative of \sqrt[3]{t^2}-4sqrt(t^3) ?
The first derivative of \sqrt[3]{t^2}-4sqrt(t^3) is (2t)/(3(t^2)^{2/3)}-6sqrt(t)