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

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

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

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

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Popular >

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to - 1 lim (\frac{x}{5 sin ( x + 1 )})

\frac{d y}{d x}, x - y = x y

\int \frac{x ^{2} + 2}{x} d x

x \to 2 lim (x + 5)

\int_{0}^{1} \frac{1}{x ^{\frac{18}{19}}} d x

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