What is the derivative of-8sin(4x) ?

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

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

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

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\frac{d}{d x} (- 8 sin (4 x))

- 32 cos (4 x)

Solution steps

\frac{d}{d x} (- 8 sin (4 x))

Take the constant out:

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

= - 8 \frac{d}{d x} (sin (4 x))

Apply the chain rule:

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

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

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

= - 8 cos (4 x) \cdot 4

Simplify

= - 32 cos (4 x)

x \to 3 + lim (x)

\int_{0}^{2} x^{2} e^{x^{3}} d x

x \to 1 lim (\frac{x ^{2} - 1}{x - 1})

\int \frac{y}{3} d y

\int \frac{e ^{x} x}{x ^{2}} d x

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