derivative of (1-cos(x)/(1+cos(x)))

1) Let's differentiate:

\frac{d}{d x} (1 - \frac{\cos (x)}{1 + \cos (x)})

ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (1 - \frac{\cos (x)}{1 + \cos (x)})

0 + \frac{d}{d x} (- \frac{\cos (x)}{1 + \cos (x)})

0 + \frac{d}{d x} (- \frac{\cos (x)}{1 + \cos (x)})

We differentiate

p1

\cos (x)

and

q2

1 + \cos (x)

using:

\frac{d}{d x} (\frac{p1}{q2}) = \frac{\frac{d}{d x} (p1) \cdot q2 - p1 \cdot \frac{d}{d x} (q2)}{{q2}^{2}}

2) Let's differentiate

p1

\frac{d}{d x} (\cos (x))

ⓘ The derivative of cos(x) is -sin(x):

\frac{d}{d x} (\cos (x))

- \sin (x)

- \sin (x)

3) Let's perform the differentiation

q2

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

ⓘ The derivative of a sum is the sum of the derivatives:

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

0 + \frac{d}{d x} (\cos (x))

0 + \frac{d}{d x} (\cos (x))

ⓘ Adding 0 changes nothing:

0 + \frac{d}{d x} (\cos (x))

\frac{d}{d x} (\cos (x))

■The derivative of cos(x) is -sin(x):

\frac{d}{d x} (\cos (x))

- \sin (x)

- \sin (x)

4) Let's evaluate the resulting derivative (

\frac{p1}{q2}

)':

- (\frac{- \sin (x) \cdot (1 + \cos (x)) - \cos (x) \cdot (- \sin (x))}{{(1 + \cos (x))}^{2}})

ⓘ Extract the minus signs: an even number of minus signs cancels out, an odd number leaves one minus sign:

- \cos (x) \cdot (- \sin (x))

\cos (x) \cdot \sin (x)

- (\frac{- \sin (x) \cdot (1 + \cos (x)) + \cos (x) \cdot \sin (x)}{{(1 + \cos (x))}^{2}})

ⓘ The minus sign of the fraction goes into the numerator or the denominator, changing the sign of each of its terms:

- (\frac{- \sin (x) \cdot (1 + \cos (x)) + \cos (x) \cdot \sin (x)}{{(1 + \cos (x))}^{2}})

\frac{\sin (x) \cdot (1 + \cos (x)) - \cos (x) \cdot \sin (x)}{{(1 + \cos (x))}^{2}}

\frac{\sin (x) \cdot (1 + \cos (x)) - \cos (x) \cdot \sin (x)}{{(1 + \cos (x))}^{2}}

ⓘ What the terms have in common is factored out, so that the fraction can be simplified:

\frac{\sin (x) \cdot (1 + \cos (x)) - \cos (x) \cdot \sin (x)}{{(1 + \cos (x))}^{2}}

\frac{\sin (x) \cdot ((\cos (x) + 1) - \cos (x))}{{(1 + \cos (x))}^{2}}

\frac{\sin (x) \cdot ((\cos (x) + 1) - \cos (x))}{{(1 + \cos (x))}^{2}}

ⓘ Like terms are grouped by adding their coefficients:

\cos (x) - \cos (x)

0

\frac{\sin (x) \cdot (0 + 1)}{{(1 + \cos (x))}^{2}}

ⓘ Adding 0 changes nothing:

0 + 1

1

\frac{\sin (x)}{{(1 + \cos (x))}^{2}}

5) Then you can continue the derivation:

ⅰ

\frac{d}{d x} (- \frac{\cos (x)}{1 + \cos (x)})

\frac{\sin (x)}{{(1 + \cos (x))}^{2}}

\frac{\sin (x)}{{(1 + \cos (x))}^{2}}

▶Derivative:

\frac{\sin (x)}{{(\cos (x) + 1)}^{2}}