What is the derivative of-2sin(4x) ?

The derivative of-2sin(4x) is -8cos(4x)

What is the first derivative of-2sin(4x) ?

The first derivative of-2sin(4x) is -8cos(4x)

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\frac{d}{d x} (- 2 sin (4 x))

- 8 cos (4 x)

Solution steps

\frac{d}{d x} (- 2 sin (4 x))

Take the constant out:

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

= - 2 \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

= - 2 cos (4 x) \cdot 4

Simplify

= - 8 cos (4 x)

\int \frac{x ^{2} - 169}{x} d x

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

ln (1 + 5 x)

\frac{d y}{d x}, y = 1 3^{x}

\frac{d}{d x} (- \frac{2}{e ^{x}})

What is the derivative of-2sin(4x) ?
The derivative of-2sin(4x) is -8cos(4x)
What is the first derivative of-2sin(4x) ?
The first derivative of-2sin(4x) is -8cos(4x)