What is (\partial)/(\partial x)(3cos(2x)) ?

The answer to (\partial)/(\partial x)(3cos(2x)) is -6sin(2x)

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(\partial)/(\partial x)(3cos(2x))

\frac{\partial}{\partial x} (3 cos (2 x))

- 6 sin (2 x)

Solution steps

\frac{\partial}{\partial x} (3 cos (2 x))

Take the constant out:

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

= 3 \frac{\partial}{\partial x} (cos (2 x))

Apply the chain rule:

- sin (2 x) \frac{\partial}{\partial x} (2 x)

= - sin (2 x) \frac{\partial}{\partial x} (2 x)

\frac{\partial}{\partial x} (2 x) = 2

= 3 (- sin (2 x) \cdot 2)

Simplify

= - 6 sin (2 x)

h (x) = \frac{1}{x ^{2}} + \frac{1}{x} + x

\int \frac{cos ( a x )}{b - a sin ( a x )} d x

x \to 5 lim (3 x + 6)

\frac{d}{d x} (arcsin (\frac{x}{1 - x}))

\int \frac{x}{x ^{2}} d x

What is (\partial)/(\partial x)(3cos(2x)) ?
The answer to (\partial)/(\partial x)(3cos(2x)) is -6sin(2x)