What is the d/(dθ)(6cos(θ)cos(θ)) ?

The d/(dθ)(6cos(θ)cos(θ)) is -12cos(θ)sin(θ)

What is the first d/(dθ)(6cos(θ)cos(θ)) ?

The first d/(dθ)(6cos(θ)cos(θ)) is -12cos(θ)sin(θ)

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\frac{d}{d θ} (6 cos (θ) cos (θ))

- 12 cos (θ) sin (θ)

Solution steps

\frac{d}{d θ} (6 cos (θ) cos (θ))

Simplify

6 cos (θ) cos (θ) : 6 cos^{2} (θ)

= \frac{d}{d θ} (6 cos^{2} (θ))

Take the constant out:

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

= 6 \frac{d}{d θ} (cos^{2} (θ))

Apply the chain rule:

2 cos (θ) \frac{d}{d θ} (cos (θ))

= 2 cos (θ) \frac{d}{d θ} (cos (θ))

\frac{d}{d θ} (cos (θ)) = - sin (θ)

= 6 \cdot 2 cos (θ) (- sin (θ))

Simplify

= - 12 cos (θ) sin (θ)

n = 1 \sum \infty \frac{4}{n ( n + 1 )}

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

x \to 0 lim (1 + 2 x)

\frac{\partial}{\partial x} (x e^{x^{2} y^{2}})

What is the d/(dθ)(6cos(θ)cos(θ)) ?
The d/(dθ)(6cos(θ)cos(θ)) is -12cos(θ)sin(θ)
What is the first d/(dθ)(6cos(θ)cos(θ)) ?
The first d/(dθ)(6cos(θ)cos(θ)) is -12cos(θ)sin(θ)