What is the derivative of-1/(3\sqrt[5]{7cos(2u))} ?

The derivative of-1/(3\sqrt[5]{7cos(2u))} is -(2sin(2u))/(15\sqrt[5]{7)cos^{6/5}(2u)}

What is the first derivative of-1/(3\sqrt[5]{7cos(2u))} ?

The first derivative of-1/(3\sqrt[5]{7cos(2u))} is -(2sin(2u))/(15\sqrt[5]{7)cos^{6/5}(2u)}

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Solution steps

Simplify

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

Apply exponent rule:

\frac{1}{a} = a^{- 1}

Apply the chain rule:

- \frac{1}{5 ( cos ( 2 u ) ) ^{\frac{6}{5}}} \frac{d}{d u} (cos (2 u))

= - \frac{1}{5 ( cos ( 2 u ) ) ^{\frac{6}{5}}} \frac{d}{d u} (cos (2 u))

\frac{d}{d u} (cos (2 u)) = - sin (2 u) \cdot 2

Simplify

n = 1 \sum \infty 3 (\frac{2}{3})^{n - 1}

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

\frac{d}{d x} (\frac{x ^{2} - 6 x + 7}{x ^{2} + 9})

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

f (x) = x^{2} + 2, - 3, 1

What is the derivative of-1/(3\sqrt[5]{7cos(2u))} ?
The derivative of-1/(3\sqrt[5]{7cos(2u))} is -(2sin(2u))/(15\sqrt[5]{7)cos^{6/5}(2u)}
What is the first derivative of-1/(3\sqrt[5]{7cos(2u))} ?
The first derivative of-1/(3\sqrt[5]{7cos(2u))} is -(2sin(2u))/(15\sqrt[5]{7)cos^{6/5}(2u)}