What is the d/(dθ)(4sin(θ)cos^2(θ)) ?

The d/(dθ)(4sin(θ)cos^2(θ)) is 4(-2cos(θ)+3cos^3(θ))

What is the first d/(dθ)(4sin(θ)cos^2(θ)) ?

The first d/(dθ)(4sin(θ)cos^2(θ)) is 4(-2cos(θ)+3cos^3(θ))

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\frac{d}{d θ} (4 sin (θ) cos^{2} (θ))

4 (- 2 cos (θ) + 3 cos^{3} (θ))

Solution steps

\frac{d}{d θ} (4 sin (θ) cos^{2} (θ))

Take the constant out:

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

= 4 \frac{d}{d θ} (sin (θ) cos^{2} (θ))

Apply the Product Rule:

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

f = sin (θ), g = cos^{2} (θ)

= 4 (\frac{d}{d θ} (sin (θ)) cos^{2} (θ) + \frac{d}{d θ} (cos^{2} (θ)) sin (θ))

\frac{d}{d θ} (sin (θ)) = cos (θ)

\frac{d}{d θ} (cos^{2} (θ)) = - sin (2 θ)

= 4 (cos (θ) cos^{2} (θ) + (- sin (2 θ)) sin (θ))

Simplify

4 (cos (θ) cos^{2} (θ) + (- sin (2 θ)) sin (θ)) : 4 (- 2 cos (θ) + 3 cos^{3} (θ))

= 4 (- 2 cos (θ) + 3 cos^{3} (θ))

3 x, 4 x^{2}

\frac{d}{d x} (a cos (4 x) + b sin (4 x))

\int 4 cos^{2} (x) sin (x) d x

\int_{0}^{4} (\frac{1}{4 - x}) d x

\frac{\partial}{\partial y} (e^{x s i n (y)})

What is the d/(dθ)(4sin(θ)cos^2(θ)) ?
The d/(dθ)(4sin(θ)cos^2(θ)) is 4(-2cos(θ)+3cos^3(θ))
What is the first d/(dθ)(4sin(θ)cos^2(θ)) ?
The first d/(dθ)(4sin(θ)cos^2(θ)) is 4(-2cos(θ)+3cos^3(θ))