What is the derivative of-8cos(-4x-3) ?

The derivative of-8cos(-4x-3) is 32sin(3+4x)

What is the first derivative of-8cos(-4x-3) ?

The first derivative of-8cos(-4x-3) is 32sin(3+4x)

\frac{d}{d x} (- 8 cos (- 4 x - 3))

32 sin (3 + 4 x)

Solution steps

\frac{d}{d x} (- 8 cos (- 4 x - 3))

Simplify

- 8 cos (- 4 x - 3) : - 8 cos (3 + 4 x)

= \frac{d}{d x} (- 8 cos (3 + 4 x))

Take the constant out:

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

= - 8 \frac{d}{d x} (cos (3 + 4 x))

Apply the chain rule:

- sin (3 + 4 x) \frac{d}{d x} (3 + 4 x)

= - sin (3 + 4 x) \frac{d}{d x} (3 + 4 x)

\frac{d}{d x} (3 + 4 x) = 4

= - 8 (- sin (3 + 4 x) \cdot 4)

Simplify

= 32 sin (3 + 4 x)

in v erse l a pl a ce \frac{1}{s ( s + 2 ) ^{2}}

\int \frac{x ^{2} + 4}{3 x ^{3} - 4 x ^{2} - 4 x} d x

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} - 3 y = 0

\int \frac{1}{y ^{2} - 2} d y

l a pl a ce t r an s f or m 2 (t - 2)^{2} + 6 (t - 2) + 6

What is the derivative of-8cos(-4x-3) ?
The derivative of-8cos(-4x-3) is 32sin(3+4x)
What is the first derivative of-8cos(-4x-3) ?
The first derivative of-8cos(-4x-3) is 32sin(3+4x)