What is (\partial)/(\partial x)(6cos^4(3x+2)) ?

The answer to (\partial)/(\partial x)(6cos^4(3x+2)) is -72cos^3(3x+2)sin(3x+2)

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(\partial)/(\partial x)(6cos^4(3x+2))

\frac{\partial}{\partial x} (6 cos^{4} (3 x + 2))

- 72 cos^{3} (3 x + 2) sin (3 x + 2)

Solution steps

\frac{\partial}{\partial x} (6 cos^{4} (3 x + 2))

Take the constant out:

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

= 6 \frac{\partial}{\partial x} (cos^{4} (3 x + 2))

Apply the chain rule:

4 (cos (3 x + 2))^{3} \frac{\partial}{\partial x} (cos (3 x + 2))

= 4 (cos (3 x + 2))^{3} \frac{\partial}{\partial x} (cos (3 x + 2))

\frac{\partial}{\partial x} (cos (3 x + 2)) = - sin (3 x + 2) \cdot 3

= 6 \cdot 4 (cos (3 x + 2))^{3} (- sin (3 x + 2) \cdot 3)

Simplify

= - 72 cos^{3} (3 x + 2) sin (3 x + 2)

\frac{d}{d x} (e^{2 x} + e^{- 2 x})

(x - 3)^{^{'}}

\frac{d}{d x} (x e^{\frac{x ^{2}}{2}})

\int \frac{3 + 4 x}{1 + x ^{2}} d x

\int_{0}^{\frac{π}{4}} x^{2} cos (2 x) d x

What is (\partial)/(\partial x)(6cos^4(3x+2)) ?
The answer to (\partial)/(\partial x)(6cos^4(3x+2)) is -72cos^3(3x+2)sin(3x+2)