What is the d/(dθ)(3-3sin(θ)) ?

The d/(dθ)(3-3sin(θ)) is -3cos(θ)

What is the first d/(dθ)(3-3sin(θ)) ?

The first d/(dθ)(3-3sin(θ)) is -3cos(θ)

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\frac{d}{d θ} (3 - 3 sin (θ))

- 3 cos (θ)

Solution steps

\frac{d}{d θ} (3 - 3 sin (θ))

Apply the Sum/Difference Rule:

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

= \frac{d}{d θ} (3) - \frac{d}{d θ} (3 sin (θ))

\frac{d}{d θ} (3) = 0

\frac{d}{d θ} (3 sin (θ)) = 3 cos (θ)

= 0 - 3 cos (θ)

Simplify

= - 3 cos (θ)

\int_{1}^{2} (\frac{1}{x ^{2}}) d x

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

\frac{\partial}{\partial x} (e^{- y} sin (x))

\frac{d y}{d x} = \frac{2 x ^{3}}{y ^{2}}, y (2) = 3

\frac{d}{d x} (17 - x^{2} - 4 y^{2})