What is the derivative of 2\sqrt[5]{x^3} ?

The derivative of 2\sqrt[5]{x^3} is (6x^2)/(5(x^3)^{4/5)}

What is the first derivative of 2\sqrt[5]{x^3} ?

The first derivative of 2\sqrt[5]{x^3} is (6x^2)/(5(x^3)^{4/5)}

derivative of

2 5 x^{3}

\frac{6 x ^{2}}{5 ( x ^{3} ) ^{\frac{4}{5}}}

Solution steps

\frac{d}{d x} (2 5 x^{3})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (5 x^{3})

Apply the chain rule:

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

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

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

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

Simplify

2 \cdot \frac{1}{5 ( x ^{3} ) ^{\frac{4}{5}}} \cdot 3 x^{2} : \frac{6 x ^{2}}{5 ( x ^{3} ) ^{\frac{4}{5}}}

= \frac{6 x ^{2}}{5 ( x ^{3} ) ^{\frac{4}{5}}}

t an g e n t f (x) = \frac{- 8 x}{x ^{2} + 1}, at x = 0

\int \frac{1}{x} - \frac{3}{x ^{2} + 1} d x

\int (2 sin (x)) d x

\frac{d}{d x} (\frac{3 x ^{3}}{2})

(y e^{x y} cos (2 x) - 2 e^{x y} sin (2 x) + 2 x) + (x e^{x y} cos (2 x) - 3) y^{^{'}} = 0

What is the derivative of 2\sqrt[5]{x^3} ?
The derivative of 2\sqrt[5]{x^3} is (6x^2)/(5(x^3)^{4/5)}
What is the first derivative of 2\sqrt[5]{x^3} ?
The first derivative of 2\sqrt[5]{x^3} is (6x^2)/(5(x^3)^{4/5)}