What is the d/(dθ)(2sin(θ)+2cos(θ)) ?

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

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

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

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\frac{d}{d θ} (2 sin (θ) + 2 cos (θ))

2 cos (θ) - 2 sin (θ)

Solution steps

\frac{d}{d θ} (2 sin (θ) + 2 cos (θ))

Apply the Sum/Difference Rule:

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

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

\frac{d}{d θ} (2 sin (θ)) = 2 cos (θ)

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

= 2 cos (θ) - 2 sin (θ)

(9.5) (21.1)

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

\frac{d}{d x} (\frac{- sin ( x )}{2})

\int \frac{2 x - 1}{1 - x ^{2}} d x

x \to + 4 + lim (2 x - 7)

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