What is the d/(dθ)(sin(θ)(4cos(θ)-4)) ?

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

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

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

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\frac{d}{d θ} (sin (θ) (4 cos (θ) - 4))

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

Solution steps

\frac{d}{d θ} (sin (θ) (4 cos (θ) - 4))

Apply the Product Rule:

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

f = sin (θ), g = 4 cos (θ) - 4

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

\frac{d}{d θ} (sin (θ)) = cos (θ)

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

= cos (θ) (4 cos (θ) - 4) + (- 4 sin (θ)) sin (θ)

Simplify

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

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

x \to 0 lim (ln^{2} (x))

\int_{0}^{5} \frac{x ^{3}}{x ^{4} + 4} d x

\frac{d}{d x} ((x - 8)^{2})

3 x^{2}, \frac{1}{3} x, 0, 0.5

f (x) = \frac{x ^{2} - 6 x}{x + 1}

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