What is the d/(dθ)(3-sin(θ)) ?

The d/(dθ)(3-sin(θ)) is -cos(θ)

What is the first d/(dθ)(3-sin(θ)) ?

The first d/(dθ)(3-sin(θ)) is -cos(θ)

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\frac{d}{d θ} (3 - sin (θ))

- cos (θ)

Solution steps

\frac{d}{d θ} (3 - sin (θ))

Apply the Sum/Difference Rule:

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

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

\frac{d}{d θ} (3) = 0

\frac{d}{d θ} (sin (θ)) = cos (θ)

= 0 - cos (θ)

Simplify

= - cos (θ)

f (x) = x^{2} - 4, at x = 0

\frac{d}{d x} (\frac{3 x ^{4} + 1}{x ^{3}})

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

\int \frac{ln ( x )}{x} d x

x^{2} \frac{d ^{2} y}{d x ^{2}} - 5 x y^{^{'}} + 8 y = 4 x^{2}