What is the derivative of sin(8-9x) ?

The derivative of sin(8-9x) is -9cos(8-9x)

What is the first derivative of sin(8-9x) ?

The first derivative of sin(8-9x) is -9cos(8-9x)

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\frac{d}{d x} (sin (8 - 9 x))

- 9 cos (8 - 9 x)

Solution steps

\frac{d}{d x} (sin (8 - 9 x))

Apply the chain rule:

cos (8 - 9 x) \frac{d}{d x} (8 - 9 x)

= cos (8 - 9 x) \frac{d}{d x} (8 - 9 x)

\frac{d}{d x} (8 - 9 x) = - 9

= cos (8 - 9 x) (- 9)

Simplify

= - 9 cos (8 - 9 x)

y = \frac{4}{7 x ^{7}}

\frac{d v}{d t} = 9.8 - \frac{v}{5}

\frac{s + 1}{s ( s ^{2} + 2 s + 2 )}

1 - t e^{- 2 t}

\int \frac{1}{4 x ^{2} + 48 x + 148} d x

What is the derivative of sin(8-9x) ?
The derivative of sin(8-9x) is -9cos(8-9x)
What is the first derivative of sin(8-9x) ?
The first derivative of sin(8-9x) is -9cos(8-9x)