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

The derivative of x\sqrt[3]{1+x^2} is cube root of 1+x^2}+(2x^2)/(3(1+x^2)^{2/3)

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

The first derivative of x\sqrt[3]{1+x^2} is cube root of 1+x^2}+(2x^2)/(3(1+x^2)^{2/3)

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Solution steps

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

\frac{d x}{d x} = 1

Simplify

\frac{d}{d x} (\frac{a x}{b})

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

y - 1 + x y^{^{'}} = 4 x

- 5 x^{2}

\frac{\partial}{\partial x} (\frac{( x ^{2} )}{y + 1})

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