What is the d/(dθ)(8/(2+cot(θ)sin(θ))) ?

The d/(dθ)(8/(2+cot(θ)sin(θ))) is (8sin(θ))/((2+cos(θ))^2)

What is the first d/(dθ)(8/(2+cot(θ)sin(θ))) ?

The first d/(dθ)(8/(2+cot(θ)sin(θ))) is (8sin(θ))/((2+cos(θ))^2)

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\frac{d}{d θ} (\frac{8}{2 + cot ( θ ) sin ( θ )})

\frac{8 sin ( θ )}{( 2 + cos ( θ ) ) ^{2}}

Solution steps

\frac{d}{d θ} (\frac{8}{2 + cot ( θ ) sin ( θ )})

Take the constant out:

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

= 8 \frac{d}{d θ} (\frac{1}{2 + cot ( θ ) sin ( θ )})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= 8 \frac{d}{d θ} ((2 + cot (θ) sin (θ))^{- 1})

Apply the chain rule:

- \frac{1}{( 2 + cot ( θ ) sin ( θ ) ) ^{2}} \frac{d}{d θ} (2 + cot (θ) sin (θ))

= - \frac{1}{( 2 + cot ( θ ) sin ( θ ) ) ^{2}} \frac{d}{d θ} (2 + cot (θ) sin (θ))

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

= 8 (- \frac{1}{( 2 + cot ( θ ) sin ( θ ) ) ^{2}} (- sin (θ)))

Simplify

8 (- \frac{1}{( 2 + cot ( θ ) sin ( θ ) ) ^{2}} (- sin (θ))) : \frac{8 sin ( θ )}{( 2 + cos ( θ ) ) ^{2}}

= \frac{8 sin ( θ )}{( 2 + cos ( θ ) ) ^{2}}

\int \frac{1}{x ( ln ( x ) ) ^{4}} d x

\int \frac{x - 2}{( 2 x + 1 ) ( x + 3 )} d x

y = (7 x + 3) x^{2}

cos (x) (\frac{π}{2})

\frac{d}{d x} (5 ln ∣ x - e^{x} ∣)

What is the d/(dθ)(8/(2+cot(θ)sin(θ))) ?
The d/(dθ)(8/(2+cot(θ)sin(θ))) is (8sin(θ))/((2+cos(θ))^2)
What is the first d/(dθ)(8/(2+cot(θ)sin(θ))) ?
The first d/(dθ)(8/(2+cot(θ)sin(θ))) is (8sin(θ))/((2+cos(θ))^2)