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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x (3 - x)^{3}

f (x) = x^{2} cos (\frac{1}{x})

y = \frac{x + 1}{x - 1}

n = 2 \sum \infty \frac{1}{n ln ( n )}

x \to 0 lim (\frac{4 cos ( x ) - 4}{x})

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