What is the d/(dθ)(4cos^3(θ)-3cos(θ)) ?

The d/(dθ)(4cos^3(θ)-3cos(θ)) is 12sin^3(θ)-9sin(θ)

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

The first d/(dθ)(4cos^3(θ)-3cos(θ)) is 12sin^3(θ)-9sin(θ)

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\frac{d}{d θ} (4 cos^{3} (θ) - 3 cos (θ))

12 sin^{3} (θ) - 9 sin (θ)

Solution steps

\frac{d}{d θ} (4 cos^{3} (θ) - 3 cos (θ))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d θ} (4 cos^{3} (θ)) - \frac{d}{d θ} (3 cos (θ))

\frac{d}{d θ} (4 cos^{3} (θ)) = - 12 cos^{2} (θ) sin (θ)

\frac{d}{d θ} (3 cos (θ)) = - 3 sin (θ)

= - 12 cos^{2} (θ) sin (θ) - (- 3 sin (θ))

Simplify

- 12 cos^{2} (θ) sin (θ) - (- 3 sin (θ)) : 12 sin^{3} (θ) - 9 sin (θ)

= 12 sin^{3} (θ) - 9 sin (θ)

f (x) = 2 x^{2} + 6 x

y = 2 (x - 3)^{2} - 2

y = - x^{2} + 2 x + 4

f (x) = 2 x^{2} - 4 x + 4

f (x) = 2 x^{2} - 4 x + 9

What is the d/(dθ)(4cos^3(θ)-3cos(θ)) ?
The d/(dθ)(4cos^3(θ)-3cos(θ)) is 12sin^3(θ)-9sin(θ)
What is the first d/(dθ)(4cos^3(θ)-3cos(θ)) ?
The first d/(dθ)(4cos^3(θ)-3cos(θ)) is 12sin^3(θ)-9sin(θ)