What is the derivative of 8sin(8x-1) ?

The derivative of 8sin(8x-1) is 64cos(8x-1)

What is the first derivative of 8sin(8x-1) ?

The first derivative of 8sin(8x-1) is 64cos(8x-1)

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derivative of

8 sin (8 x - 1)

64 cos (8 x - 1)

Solution steps

\frac{d}{d x} (8 sin (8 x - 1))

Take the constant out:

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

= 8 \frac{d}{d x} (sin (8 x - 1))

Apply the chain rule:

cos (8 x - 1) \frac{d}{d x} (8 x - 1)

= cos (8 x - 1) \frac{d}{d x} (8 x - 1)

\frac{d}{d x} (8 x - 1) = 8

= 8 cos (8 x - 1) \cdot 8

Simplify

= 64 cos (8 x - 1)

\frac{d}{d x} (e^{x} (2 x + 3)^{3})

\frac{d ^{2}}{d x ^{2}} (e^{\frac{3}{x}})

\int \frac{9 x}{x ^{2} + 1} d x

f (x) = 6 x, at x = 1

(16 y^{2} - x y) d x + x^{2} d y = 0

What is the derivative of 8sin(8x-1) ?
The derivative of 8sin(8x-1) is 64cos(8x-1)
What is the first derivative of 8sin(8x-1) ?
The first derivative of 8sin(8x-1) is 64cos(8x-1)