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

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

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

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

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Solution steps

Apply the Product Rule:

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

- \frac{3}{x}

\int \frac{x + 4}{( x ^{2} + 8 x - 7 ) ^{2}} d x

\frac{d}{d t} (t + 2)

y = tan (x^{2})

\int (3 x - 2)^{5} d x

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