What is the derivative of 2sin(4x) ?

The derivative of 2sin(4x) is 8cos(4x)

What is the first derivative of 2sin(4x) ?

The first derivative of 2sin(4x) is 8cos(4x)

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derivative of

2 sin (4 x)

8 cos (4 x)

Solution steps

\frac{d}{d x} (2 sin (4 x))

Take the constant out:

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

= 2 \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

= 2 cos (4 x) \cdot 4

Simplify

= 8 cos (4 x)

\int sin^{3} (3 x) cos^{5} (3 x) d x

\int_{- 2}^{14} \frac{1}{( x + 2 ) ^{\frac{1}{4}}} d x

\int \frac{x ^{3}}{x} d x

\int x + 2 - x^{2} d x

y^{^{'}} = 9 x + y, y (0) = 4

What is the derivative of 2sin(4x) ?
The derivative of 2sin(4x) is 8cos(4x)
What is the first derivative of 2sin(4x) ?
The first derivative of 2sin(4x) is 8cos(4x)