What is the d/(dθ)(2cos(θ)+4sin(θ)) ?

The d/(dθ)(2cos(θ)+4sin(θ)) is -2sin(θ)+4cos(θ)

What is the first d/(dθ)(2cos(θ)+4sin(θ)) ?

The first d/(dθ)(2cos(θ)+4sin(θ)) is -2sin(θ)+4cos(θ)

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\frac{d}{d θ} (2 cos (θ) + 4 sin (θ))

- 2 sin (θ) + 4 cos (θ)

Solution steps

\frac{d}{d θ} (2 cos (θ) + 4 sin (θ))

Apply the Sum/Difference Rule:

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

= \frac{d}{d θ} (2 cos (θ)) + \frac{d}{d θ} (4 sin (θ))

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

\frac{d}{d θ} (4 sin (θ)) = 4 cos (θ)

= - 2 sin (θ) + 4 cos (θ)

\frac{d}{d x} ((sin (x))^{4} + sin (x^{4}))

\frac{1}{4 s + 2}

x \to - 3 lim (- x^{2} + 9 x - 1)

\int \frac{7}{x ^{2} + 16} d x

\frac{d}{d θ} (2 θ)

What is the d/(dθ)(2cos(θ)+4sin(θ)) ?
The d/(dθ)(2cos(θ)+4sin(θ)) is -2sin(θ)+4cos(θ)
What is the first d/(dθ)(2cos(θ)+4sin(θ)) ?
The first d/(dθ)(2cos(θ)+4sin(θ)) is -2sin(θ)+4cos(θ)