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

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

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

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

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int_{0}^{3} 4 x d x

x \to - 1 lim (\frac{x ^{2} + x}{x + 1})

(2 x + y) d x + (x - 2 y) d y = 0

(ln (e^{x}))^{^{'}}

f (x) = 4 x^{2} ln (4 x)

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