What is the d/(dθ)((1+2sin(θ))cos(θ)) ?

The d/(dθ)((1+2sin(θ))cos(θ)) is 2cos^2(θ)-sin(θ)(1+2sin(θ))

What is the first d/(dθ)((1+2sin(θ))cos(θ)) ?

The first d/(dθ)((1+2sin(θ))cos(θ)) is 2cos^2(θ)-sin(θ)(1+2sin(θ))

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\frac{d}{d θ} ((1 + 2 sin (θ)) cos (θ))

2 cos^{2} (θ) - sin (θ) (1 + 2 sin (θ))

Solution steps

\frac{d}{d θ} ((1 + 2 sin (θ)) cos (θ))

Apply the Product Rule:

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

f = 1 + 2 sin (θ), g = cos (θ)

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

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

\frac{d}{d θ} (cos (θ)) = - sin (θ)

= 2 cos (θ) cos (θ) + (- sin (θ)) (1 + 2 sin (θ))

Simplify

2 cos (θ) cos (θ) + (- sin (θ)) (1 + 2 sin (θ)) : 2 cos^{2} (θ) - sin (θ) (1 + 2 sin (θ))

= 2 cos^{2} (θ) - sin (θ) (1 + 2 sin (θ))

d er i v a t i v e - 7 x^{5} - 1

\int \frac{1}{x ^{2} - 4 x + 13} d x

\frac{\partial}{\partial x} (4 x^{8 y})

\int x 2 - 1 - x^{2} d x

\int (arccos (x))^{2} d x

What is the d/(dθ)((1+2sin(θ))cos(θ)) ?
The d/(dθ)((1+2sin(θ))cos(θ)) is 2cos^2(θ)-sin(θ)(1+2sin(θ))
What is the first d/(dθ)((1+2sin(θ))cos(θ)) ?
The first d/(dθ)((1+2sin(θ))cos(θ)) is 2cos^2(θ)-sin(θ)(1+2sin(θ))