What is the derivative of f(x)=9sin(x)cos(x) ?

The derivative of f(x)=9sin(x)cos(x) is 9cos(2x)

What is the first derivative of f(x)=9sin(x)cos(x) ?

The first derivative of f(x)=9sin(x)cos(x) is 9cos(2x)

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derivative of

f (x) = 9 sin (x) cos (x)

9 cos (2 x)

Solution steps

\frac{d}{d x} (9 sin (x) cos (x))

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

= 9 (cos (x) cos (x) + (- sin (x)) sin (x))

Simplify

9 (cos (x) cos (x) + (- sin (x)) sin (x)) : 9 cos (2 x)

= 9 cos (2 x)

\int sin (x) cos (2 x) d x

cos (x^{3})

\int_{1}^{\infty} 5 e^{- 5 x} d x

\int e^{- x^{2}} d x

What is the derivative of f(x)=9sin(x)cos(x) ?
The derivative of f(x)=9sin(x)cos(x) is 9cos(2x)
What is the first derivative of f(x)=9sin(x)cos(x) ?
The first derivative of f(x)=9sin(x)cos(x) is 9cos(2x)