What is the d/(dθ)(2cos(θ)) ?

The d/(dθ)(2cos(θ)) is -2sin(θ)

What is the first d/(dθ)(2cos(θ)) ?

The first d/(dθ)(2cos(θ)) is -2sin(θ)

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\frac{d}{d θ} (2 cos (θ))

- 2 sin (θ)

Solution steps

\frac{d}{d θ} (2 cos (θ))

Take the constant out:

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

= 2 \frac{d}{d θ} (cos (θ))

Apply the common derivative:

\frac{d}{d θ} (cos (θ)) = - sin (θ)

= 2 (- sin (θ))

Simplify

= - 2 sin (θ)

\frac{d y}{d x} = (4 x^{2} + 3 x + 2) (y + 5)

\int 1.5 x d x

(x^{2} + 4), (- x^{2} - 1), x = - 2, x = 3

t \to 2 lim (6 t^{2} - 3 t + 1)

\frac{- s - 4}{s ^{2} + 4 s + 3}