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)

f (x) = \frac{20}{( x + 3 ) ^{2}}

\frac{d}{d x} (8^{c o s (x)})

f (x) = - x^{2} + 4 x, at x = 3

\int \frac{sin ( x )}{sin ( x ) + 1} d x

\frac{d X}{d X}