What is the derivative of f(x)=\sqrt[3]{x^7-4x} ?

The derivative of f(x)=\sqrt[3]{x^7-4x} is (7x^6-4)/(3(x^7-4x)^{2/3)}

What is the first derivative of f(x)=\sqrt[3]{x^7-4x} ?

The first derivative of f(x)=\sqrt[3]{x^7-4x} is (7x^6-4)/(3(x^7-4x)^{2/3)}

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derivative of

\frac{7 x ^{6} - 4}{3 ( x ^{7} - 4 x ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( x ^{7} - 4 x ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{7} - 4 x)

= \frac{1}{3 ( x ^{7} - 4 x ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{7} - 4 x)

\frac{d}{d x} (x^{7} - 4 x) = 7 x^{6} - 4

= \frac{1}{3 ( x ^{7} - 4 x ) ^{\frac{2}{3}}} (7 x^{6} - 4)

Simplify

\frac{1}{3 ( x ^{7} - 4 x ) ^{\frac{2}{3}}} (7 x^{6} - 4) : \frac{7 x ^{6} - 4}{3 ( x ^{7} - 4 x ) ^{\frac{2}{3}}}

= \frac{7 x ^{6} - 4}{3 ( x ^{7} - 4 x ) ^{\frac{2}{3}}}

\frac{d}{d x} (sin (3 x) + cos (4 x))

\frac{d}{d t} (r (t) cos (t))

3 x y d y = (y^{2} - x^{2}) d x

\int \frac{4}{x} - \frac{16}{x ^{2}} d x

f (x) = (4 x - 1) cos (x)

What is the derivative of f(x)=\sqrt[3]{x^7-4x} ?
The derivative of f(x)=\sqrt[3]{x^7-4x} is (7x^6-4)/(3(x^7-4x)^{2/3)}
What is the first derivative of f(x)=\sqrt[3]{x^7-4x} ?
The first derivative of f(x)=\sqrt[3]{x^7-4x} is (7x^6-4)/(3(x^7-4x)^{2/3)}