What is the derivative of-16cos(4x) ?

The derivative of-16cos(4x) is 64sin(4x)

What is the first derivative of-16cos(4x) ?

The first derivative of-16cos(4x) is 64sin(4x)

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\frac{d}{d x} (- 16 cos (4 x))

64 sin (4 x)

Solution steps

\frac{d}{d x} (- 16 cos (4 x))

Take the constant out:

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

= - 16 \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

= - 16 (- sin (4 x) \cdot 4)

Simplify

= 64 sin (4 x)

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

\int \frac{25 + x ^{2}}{x} d x

\frac{\partial}{\partial y} (x 1 + y^{7})

f (x) = \frac{6 x}{x + 2}, at x = 4

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x}

What is the derivative of-16cos(4x) ?
The derivative of-16cos(4x) is 64sin(4x)
What is the first derivative of-16cos(4x) ?
The first derivative of-16cos(4x) is 64sin(4x)