What is the derivative of-4cos(4x) ?

The derivative of-4cos(4x) is 16sin(4x)

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

The first derivative of-4cos(4x) is 16sin(4x)

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\frac{d}{d x} (- 4 cos (4 x))

16 sin (4 x)

Solution steps

\frac{d}{d x} (- 4 cos (4 x))

Take the constant out:

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

= - 4 \frac{d}{d x} (cos (4 x))

Apply the chain rule:

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

= - sin (4 x) \frac{d}{d x} (4 x)

\frac{d}{d x} (4 x) = 4

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

Simplify

= 16 sin (4 x)

\frac{\partial}{\partial y} (sin (x) + π cos (y))

2 v^{'} t (x) = v

\int \frac{( 1 - y ) e ^{y}}{y ^{2}} d y

y (8 x - 9 y) d x + 2 x (x - 3 y) d y = 0

\frac{d}{d x} (a x \cdot e^{b x})

What is the derivative of-4cos(4x) ?
The derivative of-4cos(4x) is 16sin(4x)
What is the first derivative of-4cos(4x) ?
The first derivative of-4cos(4x) is 16sin(4x)