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 (e^{(2x)/3}x^2(2x+3))/(3(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 (e^{(2x)/3}x^2(2x+3))/(3(x^3)^{2/3)}

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derivative of

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

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

Simplify

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

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

\int sin^{2} (n x) d x

y = \frac{( x + b ) ( x - a )}{( x - b )}

\int e^{(x - y)} d y

\int t \cdot sec^{2} (2 t) d t

y^{^{'}} - \frac{y}{t} = 0

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 (e^{(2x)/3}x^2(2x+3))/(3(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 (e^{(2x)/3}x^2(2x+3))/(3(x^3)^{2/3)}