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

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

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

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

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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 x ) ) ^{\frac{6}{5}}} \frac{d}{d x} (cos (2 x))

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

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

Simplify

\int sin^{4} (1) d x

\int \frac{x}{( x - 4 ) ^{3}} d x

2^{c o s (x)}

\frac{d}{d x} ((5 x + 3)^{4})

\frac{d}{d x} ((arctan (7 x))^{2})

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