What is the derivative of 4sin(2x-3cos(3x)) ?

The derivative of 4sin(2x-3cos(3x)) is 8cos(2x)+9sin(3x)

What is the first derivative of 4sin(2x-3cos(3x)) ?

The first derivative of 4sin(2x-3cos(3x)) is 8cos(2x)+9sin(3x)

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\frac{d}{d x} (4 sin (2 x) - 3 cos (3 x))

8 cos (2 x) + 9 sin (3 x)

Solution steps

\frac{d}{d x} (4 sin (2 x) - 3 cos (3 x))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (4 sin (2 x)) - \frac{d}{d x} (3 cos (3 x))

\frac{d}{d x} (4 sin (2 x)) = 8 cos (2 x)

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

= 8 cos (2 x) - (- 9 sin (3 x))

Simplify

= 8 cos (2 x) + 9 sin (3 x)

What is the derivative of 4sin(2x-3cos(3x)) ?
The derivative of 4sin(2x-3cos(3x)) is 8cos(2x)+9sin(3x)
What is the first derivative of 4sin(2x-3cos(3x)) ?
The first derivative of 4sin(2x-3cos(3x)) is 8cos(2x)+9sin(3x)