What is the derivative of \sqrt[6]{3+tan(t)} ?

The derivative of \sqrt[6]{3+tan(t)} is (sec^2(t))/(6(3+tan(t))^{5/6)}

What is the first derivative of \sqrt[6]{3+tan(t)} ?

The first derivative of \sqrt[6]{3+tan(t)} is (sec^2(t))/(6(3+tan(t))^{5/6)}

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derivative of

\frac{sec ^{2} ( t )}{6 ( 3 + tan ( t ) ) ^{\frac{5}{6}}}

Solution steps

Apply the chain rule:

\frac{1}{6 ( 3 + tan ( t ) ) ^{\frac{5}{6}}} \frac{d}{d t} (3 + tan (t))

= \frac{1}{6 ( 3 + tan ( t ) ) ^{\frac{5}{6}}} \frac{d}{d t} (3 + tan (t))

\frac{d}{d t} (3 + tan (t)) = sec^{2} (t)

= \frac{1}{6 ( 3 + tan ( t ) ) ^{\frac{5}{6}}} sec^{2} (t)

Simplify

\frac{1}{6 ( 3 + tan ( t ) ) ^{\frac{5}{6}}} sec^{2} (t) : \frac{sec ^{2} ( t )}{6 ( 3 + tan ( t ) ) ^{\frac{5}{6}}}

= \frac{sec ^{2} ( t )}{6 ( 3 + tan ( t ) ) ^{\frac{5}{6}}}

\frac{d}{d x} (\frac{2 x ^{5} - 15}{x ^{4}})

e^{6 y}

\frac{d}{d x} (sin (3 x) + cos (2 x))

\int_{0}^{1} \frac{2 x ^{2}}{1 + x ^{2}} d x

What is the derivative of \sqrt[6]{3+tan(t)} ?
The derivative of \sqrt[6]{3+tan(t)} is (sec^2(t))/(6(3+tan(t))^{5/6)}
What is the first derivative of \sqrt[6]{3+tan(t)} ?
The first derivative of \sqrt[6]{3+tan(t)} is (sec^2(t))/(6(3+tan(t))^{5/6)}