What is the derivative of 9sin(θ)cos(θ) ?

The derivative of 9sin(θ)cos(θ) is 9cos(2θ)

What is the first derivative of 9sin(θ)cos(θ) ?

The first derivative of 9sin(θ)cos(θ) is 9cos(2θ)

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derivative of

9 sin (θ) cos (θ)

9 cos (2 θ)

Solution steps

\frac{d}{d θ} (9 sin (θ) cos (θ))

Take the constant out:

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

= 9 \frac{d}{d θ} (sin (θ) cos (θ))

Apply the Product Rule:

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

f = sin (θ), g = cos (θ)

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

\frac{d}{d θ} (sin (θ)) = cos (θ)

\frac{d}{d θ} (cos (θ)) = - sin (θ)

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

Simplify

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

= 9 cos (2 θ)

\frac{d}{d x} (x ln (x) - 1)

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

\int \frac{( e ^{2 x} - 1 )}{( e ^{x} + e ^{- x} )} d x

\frac{d}{d x} (\frac{2 x}{3 x ^{2} - 12})

(x^{3} + 4) (3 x^{2} - 4 x + 2)

What is the derivative of 9sin(θ)cos(θ) ?
The derivative of 9sin(θ)cos(θ) is 9cos(2θ)
What is the first derivative of 9sin(θ)cos(θ) ?
The first derivative of 9sin(θ)cos(θ) is 9cos(2θ)