What is the derivative of \sqrt[3]{(x^2-1)^2} ?

The derivative of \sqrt[3]{(x^2-1)^2} is (4x(x^2-1))/(3((x^2-1)^2)^{2/3)}

What is the first derivative of \sqrt[3]{(x^2-1)^2} ?

The first derivative of \sqrt[3]{(x^2-1)^2} is (4x(x^2-1))/(3((x^2-1)^2)^{2/3)}

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derivative of

\frac{4 x ( x ^{2} - 1 )}{3 ( ( x ^{2} - 1 ) ^{2} ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

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

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

\frac{d}{d x} ((x^{2} - 1)^{2}) = 4 x (x^{2} - 1)

= \frac{1}{3 ( ( x ^{2} - 1 ) ^{2} ) ^{\frac{2}{3}}} \cdot 4 x (x^{2} - 1)

Simplify

\frac{1}{3 ( ( x ^{2} - 1 ) ^{2} ) ^{\frac{2}{3}}} \cdot 4 x (x^{2} - 1) : \frac{4 x ( x ^{2} - 1 )}{3 ( ( x ^{2} - 1 ) ^{2} ) ^{\frac{2}{3}}}

= \frac{4 x ( x ^{2} - 1 )}{3 ( ( x ^{2} - 1 ) ^{2} ) ^{\frac{2}{3}}}

\frac{d y}{d x} = \frac{3 x}{5 y}

x^{2} + 4

\frac{\partial}{\partial x} (y^{x z} + 2 (yz)^{x})

\int \frac{tan ( ln ( y ) )}{y} d y

What is the derivative of \sqrt[3]{(x^2-1)^2} ?
The derivative of \sqrt[3]{(x^2-1)^2} is (4x(x^2-1))/(3((x^2-1)^2)^{2/3)}
What is the first derivative of \sqrt[3]{(x^2-1)^2} ?
The first derivative of \sqrt[3]{(x^2-1)^2} is (4x(x^2-1))/(3((x^2-1)^2)^{2/3)}