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

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

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

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

derivative of

4 cos (sin (2 x))

- 8 sin (sin (2 x)) cos (2 x)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

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

= - sin (sin (2 x)) \frac{d}{d x} (sin (2 x))

\frac{d}{d x} (sin (2 x)) = cos (2 x) \cdot 2

= 4 (- sin (sin (2 x)) cos (2 x) \cdot 2)

Simplify

= - 8 sin (sin (2 x)) cos (2 x)

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

d er i v a t i v e e^{- x^{5}}

\int (u + 3) (8 u + 5) d u

\int \frac{e ^{4 x}}{( e ^{2 x} - 1 ) ^{3}} d x

x \to 2 + lim (\frac{1}{( x - 2 ) ^{3}})

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