What is the first (4sin(2x))^' ?

The first (4sin(2x))^' is 8cos(2x)

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(4sin(2x))^'

(4 sin (2 x))^{^{'}}

8 cos (2 x)

Solution steps

(4 sin (2 x))^{'}

Take the constant out:

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

= 4 (sin (2 x))^{^{'}}

Apply the chain rule:

cos (2 x) (2 x)^{'}

= cos (2 x) (2 x)^{'}

(2 x)^{'} = 2

= 4 cos (2 x) \cdot 2

Simplify

= 8 cos (2 x)

\int 3 x^{2} \cdot e^{x^{3}} d x

e^{y} y^{^{'}} = e^{y} + 5 x, y (3) = 7

\int \frac{1}{81 e ^{- 3 x} + e ^{3 x}} d x

\frac{d}{d x} ((cos (x)) (sin (x)))

x \to 4 lim (\frac{( 5 - x )}{x - 4})