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

The derivative of-16cos(4t) is 64sin(4t)

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

The first derivative of-16cos(4t) is 64sin(4t)

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derivative of

- 16 cos (4 t)

64 sin (4 t)

Solution steps

\frac{d}{d t} (- 16 cos (4 t))

Take the constant out:

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

= - 16 \frac{d}{d t} (cos (4 t))

Apply the chain rule:

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

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

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

= - 16 (- sin (4 t) \cdot 4)

Simplify

= 64 sin (4 t)

(3 x^{2} - 3 x) e^{x}

\frac{d}{d x} (\frac{1}{x ( 1 + ln ( x ) ) ^{2}})

n \to \infty lim ({3})

\frac{d}{d x} (\frac{- 5}{2 x ^{6} - 4 x ^{3} - 6 x})

f (x) = 2 x + \frac{4}{x}, at x = 4

What is the derivative of-16cos(4t) ?
The derivative of-16cos(4t) is 64sin(4t)
What is the first derivative of-16cos(4t) ?
The first derivative of-16cos(4t) is 64sin(4t)