What is the d/(dθ)(9sin(θ)) ?

The d/(dθ)(9sin(θ)) is 9cos(θ)

What is the first d/(dθ)(9sin(θ)) ?

The first d/(dθ)(9sin(θ)) is 9cos(θ)

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\frac{d}{d θ} (9 sin (θ))

9 cos (θ)

Solution steps

\frac{d}{d θ} (9 sin (θ))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 9 \frac{d}{d θ} (sin (θ))

Apply the common derivative:

\frac{d}{d θ} (sin (θ)) = cos (θ)

= 9 cos (θ)

\int 16 ln (9)

\frac{d}{d y} (\frac{2 x y}{( x ^{2} + y ^{2} ) ^{2}})

\frac{d}{d x} (9 x + 1)

\int_{1}^{3} \frac{2}{( x - 2 ) ^{\frac{8}{3}}} d x

2 x^{2} - 5