What is the d/(dθ)(sin(θ)+2cos(θ)) ?

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

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

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

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\frac{d}{d θ} (sin (θ) + 2 cos (θ))

cos (θ) - 2 sin (θ)

Solution steps

\frac{d}{d θ} (sin (θ) + 2 cos (θ))

Apply the Sum/Difference Rule:

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

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

\frac{d}{d θ} (sin (θ)) = cos (θ)

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

= cos (θ) - 2 sin (θ)

\int \frac{24 x ^{6} - 24 x ^{4} + x}{x} d x

\int_{- y}^{y} ∣ x ∣ d x

x d y = d x

\frac{\partial}{\partial x} (\frac{6 x}{9 + x ^{2} + y ^{2}})

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

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