What is the derivative of-8cos(2x) ?

The derivative of-8cos(2x) is 16sin(2x)

What is the first derivative of-8cos(2x) ?

The first derivative of-8cos(2x) is 16sin(2x)

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\frac{d}{d x} (- 8 cos (2 x))

16 sin (2 x)

Solution steps

\frac{d}{d x} (- 8 cos (2 x))

Take the constant out:

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

= - 8 \frac{d}{d x} (cos (2 x))

Apply the chain rule:

- sin (2 x) \frac{d}{d x} (2 x)

= - sin (2 x) \frac{d}{d x} (2 x)

\frac{d}{d x} (2 x) = 2

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

Simplify

= 16 sin (2 x)

(- 10, 4), (2, - 5)

x y^{^{'}} = 3 y

\frac{d y}{d x} = \frac{x ^{5} - 8 y}{x}

e^{x} + cos (x)

\frac{d}{d x} (sin (x) + sin (y))

What is the derivative of-8cos(2x) ?
The derivative of-8cos(2x) is 16sin(2x)
What is the first derivative of-8cos(2x) ?
The first derivative of-8cos(2x) is 16sin(2x)