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

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

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

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

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derivative of

\frac{2 e ^{2 x} + 3 x ^{2}}{3 ( e ^{2 x} + x ^{3} ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( e ^{2 x} + x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (e^{2 x} + x^{3})

= \frac{1}{3 ( e ^{2 x} + x ^{3} ) ^{\frac{2}{3}}} \frac{d}{d x} (e^{2 x} + x^{3})

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

= \frac{1}{3 ( e ^{2 x} + x ^{3} ) ^{\frac{2}{3}}} (e^{2 x} \cdot 2 + 3 x^{2})

Simplify

\frac{1}{3 ( e ^{2 x} + x ^{3} ) ^{\frac{2}{3}}} (e^{2 x} \cdot 2 + 3 x^{2}) : \frac{2 e ^{2 x} + 3 x ^{2}}{3 ( e ^{2 x} + x ^{3} ) ^{\frac{2}{3}}}

= \frac{2 e ^{2 x} + 3 x ^{2}}{3 ( e ^{2 x} + x ^{3} ) ^{\frac{2}{3}}}

f (x) = x^{\frac{5}{6}}

y = 10 - 3 x - x^{2}

\frac{d y}{d x} + y cos (x) = 4 cos (x), y (0) = 6

\int 3 x 4 - x^{2} d x

\int sin (3) (θ) cos (6) (θ) d θ

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