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

The derivative of-8sin(8x) is -64cos(8x)

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

The first derivative of-8sin(8x) is -64cos(8x)

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

- 64 cos (8 x)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= - 8 cos (8 x) \cdot 8

Simplify

= - 64 cos (8 x)

\frac{d}{d x} ((5 x^{3} + 8 x - 3)^{\frac{3}{2}})

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

3 \frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = 0

5 x - 2 x

y^{^{'}} + 3 y = 8, y (0) = 0

What is the derivative of-8sin(8x) ?
The derivative of-8sin(8x) is -64cos(8x)
What is the first derivative of-8sin(8x) ?
The first derivative of-8sin(8x) is -64cos(8x)