What is the derivative of 8cos(4x) ?

The derivative of 8cos(4x) is -32sin(4x)

What is the first derivative of 8cos(4x) ?

The first derivative of 8cos(4x) is -32sin(4x)

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derivative of

8 cos (4 x)

- 32 sin (4 x)

Solution steps

\frac{d}{d x} (8 cos (4 x))

Take the constant out:

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

= 8 \frac{d}{d x} (cos (4 x))

Apply the chain rule:

- sin (4 x) \frac{d}{d x} (4 x)

= - sin (4 x) \frac{d}{d x} (4 x)

\frac{d}{d x} (4 x) = 4

= 8 (- sin (4 x) \cdot 4)

Simplify

= - 32 sin (4 x)

\int a^{n x} d x

\frac{d}{d x} (e^{x + 6})

f (x) = ln (x^{2} + 5)

y^{^{''}} - 36 y = e^{6 x}

\frac{d}{d x} (tan (x^{2}) + tan^{2} (x))

What is the derivative of 8cos(4x) ?
The derivative of 8cos(4x) is -32sin(4x)
What is the first derivative of 8cos(4x) ?
The first derivative of 8cos(4x) is -32sin(4x)