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

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

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

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

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\frac{d}{d x} (- 16 sin (4 x))

- 64 cos (4 x)

Solution steps

\frac{d}{d x} (- 16 sin (4 x))

Take the constant out:

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

= - 16 \frac{d}{d x} (sin (4 x))

Apply the chain rule:

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

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

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

= - 16 cos (4 x) \cdot 4

Simplify

= - 64 cos (4 x)

\frac{d}{d x} (ln (\frac{1 + x}{x ^{3}}))

\int_{\frac{1}{5}}^{7} 8 x ln (5 x) d x

y^{^{'}} + \frac{2}{x} y = 4 x

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

\frac{d}{d x} (ln (7 x^{2} + 5 y^{2}))

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