What is the first (-8cos(2x))^' ?

The first (-8cos(2x))^' is 16sin(2x)

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(-8cos(2x))^'

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

16 sin (2 x)

Solution steps

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

Take the constant out:

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

= - 8 (cos (2 x))^{^{'}}

Apply the chain rule:

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

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

(2 x)^{'} = 2

= - 8 (- sin (2 x) \cdot 2)

Simplify

= 16 sin (2 x)

f (x) = 1 + x

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

\frac{\partial}{\partial x} (9 x ln (x) y)

(- 4 - 4) (- 4 - 9)

t \cdot y^{^{'}} = y + ((\frac{1}{4}) \cdot t^{2} + y^{2})^{(\frac{1}{2})}