Question

Find the derivative of sinx/1+cosx, using quotient rule​

Answer 1

Answer:

f'(x) = -1/(1 - Cos(x))

Step-by-step explanation:

The quotient rule for derivation is:

For f(x) = h(x)/k(x)

$f'(x) = \frac{h'(x)*k(x) - k'(x)*h(x)}{k^2(x)}$

In this case, the function is:

f(x) = Sin(x)/(1 + Cos(x))

Then we have:

h(x) = Sin(x)

h'(x) = Cos(x)

And for the denominator:

k(x) = 1 - Cos(x)

k'(x) = -( -Sin(x)) = Sin(x)

Replacing these in the rule, we get:

$f'(x) = \frac{Cos(x)*(1 - Cos(x)) - Sin(x)*Sin(x)}{(1 - Cos(x))^2}$

Now we can simplify that:

$f'(x) = \frac{Cos(x)*(1 - Cos(x)) - Sin(x)*Sin(x)}{(1 - Cos(x))^2} = \frac{Cos(x) - Cos^2(x) - Sin^2(x)}{(1 - Cos(x))^2}$

And we know that:

cos^2(x) + sin^2(x) = 1

then:

$f'(x) = \frac{Cos(x)- 1}{(1 - Cos(x))^2} = - \frac{(1 - Cos(x))}{(1 - Cos(x))^2} = \frac{-1}{1 - Cos(x)}$