What is the derivative of \sqrt[3]{1+tan(x}) ?

The derivative of \sqrt[3]{1+tan(x}) is (sec^2(x))/(3(1+tan(x))^{2/3)}

What is the first derivative of \sqrt[3]{1+tan(x}) ?

The first derivative of \sqrt[3]{1+tan(x}) is (sec^2(x))/(3(1+tan(x))^{2/3)}

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

y^{^{'}} - 2 x (y^{2} + 1) = 0

\frac{d}{d x} ((4 x + 1)^{2} (1 - x)^{3})

x \to 0 lim (\frac{x + 1}{x + 4})

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

y^{2} + 1 d x = x y d y

What is the derivative of \sqrt[3]{1+tan(x}) ?
The derivative of \sqrt[3]{1+tan(x}) is (sec^2(x))/(3(1+tan(x))^{2/3)}
What is the first derivative of \sqrt[3]{1+tan(x}) ?
The first derivative of \sqrt[3]{1+tan(x}) is (sec^2(x))/(3(1+tan(x))^{2/3)}