What is the derivative of (sin(3xcos(2x))/3) ?

The derivative of (sin(3xcos(2x))/3) is cos(3x)cos(2x)-2/3 sin(2x)sin(3x)

What is the first derivative of (sin(3xcos(2x))/3) ?

The first derivative of (sin(3xcos(2x))/3) is cos(3x)cos(2x)-2/3 sin(2x)sin(3x)

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

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

Solution steps

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

Take the constant out:

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

= \frac{1}{3} \frac{d}{d x} (sin (3 x) cos (2 x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = sin (3 x), g = cos (2 x)

= \frac{1}{3} (\frac{d}{d x} (sin (3 x)) cos (2 x) + \frac{d}{d x} (cos (2 x)) sin (3 x))

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

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

= \frac{1}{3} (cos (3 x) \cdot 3 cos (2 x) + (- sin (2 x) \cdot 2) sin (3 x))

Simplify

\frac{1}{3} (cos (3 x) \cdot 3 cos (2 x) + (- sin (2 x) \cdot 2) sin (3 x)) : cos (3 x) cos (2 x) - \frac{2}{3} sin (2 x) sin (3 x)

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

\int e^{x} e^{e^{x}} d x

n = 2 \sum \infty \frac{1}{n ^{2} + 6 n + 8}

- (3 x - 4)^{3}, 3

\int_{0}^{\frac{3 π}{2}} 5 ∣ sin (x) ∣ d x

\frac{x ^{2}}{2 + x}

What is the derivative of (sin(3xcos(2x))/3) ?
The derivative of (sin(3xcos(2x))/3) is cos(3x)cos(2x)-2/3 sin(2x)sin(3x)
What is the first derivative of (sin(3xcos(2x))/3) ?
The first derivative of (sin(3xcos(2x))/3) is cos(3x)cos(2x)-2/3 sin(2x)sin(3x)