What is the derivative of g(x)=sin(9-3x) ?

The derivative of g(x)=sin(9-3x) is -3cos(9-3x)

What is the first derivative of g(x)=sin(9-3x) ?

The first derivative of g(x)=sin(9-3x) is -3cos(9-3x)

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derivative of

g (x) = sin (9 - 3 x)

- 3 cos (9 - 3 x)

Solution steps

\frac{d}{d x} (sin (9 - 3 x))

Apply the chain rule:

cos (9 - 3 x) \frac{d}{d x} (9 - 3 x)

= cos (9 - 3 x) \frac{d}{d x} (9 - 3 x)

\frac{d}{d x} (9 - 3 x) = - 3

= cos (9 - 3 x) (- 3)

Simplify

= - 3 cos (9 - 3 x)

f (x) = \frac{x}{9 x ^{2} - 1}

\frac{d}{d x} (ln (\frac{x + 3}{x - 3}))

x \to 0 + lim (\frac{x}{15 x ^{4} + 12 x})

x \to 2 lim (\frac{x ^{2} - 4}{x - 3})

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

What is the derivative of g(x)=sin(9-3x) ?
The derivative of g(x)=sin(9-3x) is -3cos(9-3x)
What is the first derivative of g(x)=sin(9-3x) ?
The first derivative of g(x)=sin(9-3x) is -3cos(9-3x)