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

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

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

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

- 6 sin (3 x)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

- sin (3 x) \frac{\partial}{\partial x} (3 x)

= - sin (3 x) \frac{\partial}{\partial x} (3 x)

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

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

Simplify

= - 6 sin (3 x)

\frac{d y}{d x} + y^{5} x + 4 y = 0

\int_{0}^{2} x^{2} e^{- x} d x

\int - 2^{x} d x

(x^{3} + 1) (x^{2} + x + 1)

\frac{d}{d x} (\frac{5}{3 x ^{2}})

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