What is the d/(dθ)(sin(θ)+θcos(θ)) ?

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

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

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

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\frac{d}{d θ} (sin (θ) + θ cos (θ))

2 cos (θ) - θ sin (θ)

Solution steps

\frac{d}{d θ} (sin (θ) + θ cos (θ))

Apply the Sum/Difference Rule:

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

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

\frac{d}{d θ} (sin (θ)) = cos (θ)

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

= cos (θ) + cos (θ) - θ sin (θ)

Simplify

= 2 cos (θ) - θ sin (θ)

\frac{\partial}{\partial x} (2 sin (3 x + 2 y))

\int \frac{y + 1}{y} d y

\frac{d}{d x} (e^{x} cos (x^{2}))

y \to - 1 lim (- y^{5} - 3 y^{4})

f (x) = e^{2 x}, (0, 1)

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