What is the derivative of-4sin(4t) ?

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

What is the first derivative of-4sin(4t) ?

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

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derivative of

- 4 sin (4 t)

- 16 cos (4 t)

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

cos (4 t) \frac{d}{d t} (4 t)

= cos (4 t) \frac{d}{d t} (4 t)

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

= - 4 cos (4 t) \cdot 4

Simplify

= - 16 cos (4 t)

\frac{d}{d θ} (6 sin (θ) - 8 cos (θ))

\int 2 x^{- 2} d x

ln (4 z + 1)

x \to 1 lim (3 x^{3} - 4 x^{2} + x + 5)

\frac{d}{d x} (sin (x) + \frac{8}{7} cot (x))

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