What is the d/(dθ)((sin(θ))/(1+cos(θ))) ?

The d/(dθ)((sin(θ))/(1+cos(θ))) is 1/(1+cos(θ))

What is the first d/(dθ)((sin(θ))/(1+cos(θ))) ?

The first d/(dθ)((sin(θ))/(1+cos(θ))) is 1/(1+cos(θ))

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\frac{d}{d θ} (\frac{sin ( θ )}{1 + cos ( θ )})

\frac{1}{1 + cos ( θ )}

Solution steps

\frac{d}{d θ} (\frac{sin ( θ )}{1 + cos ( θ )})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d θ} (sin (θ)) = cos (θ)

\frac{d}{d θ} (1 + cos (θ)) = - sin (θ)

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

Simplify

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

= \frac{1}{1 + cos ( θ )}

What is the d/(dθ)((sin(θ))/(1+cos(θ))) ?
The d/(dθ)((sin(θ))/(1+cos(θ))) is 1/(1+cos(θ))
What is the first d/(dθ)((sin(θ))/(1+cos(θ))) ?
The first d/(dθ)((sin(θ))/(1+cos(θ))) is 1/(1+cos(θ))