What is the derivative of f(x)=\sqrt[5]{8+4x+x^3} ?

The derivative of f(x)=\sqrt[5]{8+4x+x^3} is (3x^2+4)/(5(8+4x+x^3)^{4/5)}

What is the first derivative of f(x)=\sqrt[5]{8+4x+x^3} ?

The first derivative of f(x)=\sqrt[5]{8+4x+x^3} is (3x^2+4)/(5(8+4x+x^3)^{4/5)}

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derivative of

\frac{3 x ^{2} + 4}{5 ( 8 + 4 x + x ^{3} ) ^{\frac{4}{5}}}

Solution steps

Apply the chain rule:

\frac{1}{5 ( 8 + 4 x + x ^{3} ) ^{\frac{4}{5}}} \frac{d}{d x} (8 + 4 x + x^{3})

= \frac{1}{5 ( 8 + 4 x + x ^{3} ) ^{\frac{4}{5}}} \frac{d}{d x} (8 + 4 x + x^{3})

\frac{d}{d x} (8 + 4 x + x^{3}) = 3 x^{2} + 4

= \frac{1}{5 ( 8 + 4 x + x ^{3} ) ^{\frac{4}{5}}} (3 x^{2} + 4)

Simplify

\frac{1}{5 ( 8 + 4 x + x ^{3} ) ^{\frac{4}{5}}} (3 x^{2} + 4) : \frac{3 x ^{2} + 4}{5 ( 8 + 4 x + x ^{3} ) ^{\frac{4}{5}}}

= \frac{3 x ^{2} + 4}{5 ( 8 + 4 x + x ^{3} ) ^{\frac{4}{5}}}

y^{^{''}} - 2 y^{^{'}} - 3 y = - 9 t e - t

\frac{\partial}{\partial y} (x^{3} + x y^{2})

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 2 e^{x}

\int 2 x + 4 x ln (8 x) d x

\frac{d}{d x} (1 + \frac{1}{3 x})

What is the derivative of f(x)=\sqrt[5]{8+4x+x^3} ?
The derivative of f(x)=\sqrt[5]{8+4x+x^3} is (3x^2+4)/(5(8+4x+x^3)^{4/5)}
What is the first derivative of f(x)=\sqrt[5]{8+4x+x^3} ?
The first derivative of f(x)=\sqrt[5]{8+4x+x^3} is (3x^2+4)/(5(8+4x+x^3)^{4/5)}