What is the derivative of 2sin(8x) ?

The derivative of 2sin(8x) is 16cos(8x)

What is the first derivative of 2sin(8x) ?

The first derivative of 2sin(8x) is 16cos(8x)

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\frac{d}{d x} (2 sin (8 x))

16 cos (8 x)

Solution steps

\frac{d}{d x} (2 sin (8 x))

Take the constant out:

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

= 2 \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

= 2 cos (8 x) \cdot 8

Simplify

= 16 cos (8 x)

x - x

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

f^{^{'}} (x) = x (6 + 5 x)

f (x) = \frac{1}{10 x}, at x = 9

\int x^{6} + \frac{1}{1 - 4 x ^{2}} d x

What is the derivative of 2sin(8x) ?
The derivative of 2sin(8x) is 16cos(8x)
What is the first derivative of 2sin(8x) ?
The first derivative of 2sin(8x) is 16cos(8x)