Question

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

Answer 1

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

Explanation:

The quotient rule for derivation is:

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

`f'(x) = (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) = (Cos(x)*(1 - Cos(x)) - Sin(x)*Sin(x))/((1 - Cos(x))^2)`

Now we can simplify that:

`f'(x) = (Cos(x)*(1 - Cos(x)) - Sin(x)*Sin(x))/((1 - Cos(x))^2) = (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) = (Cos(x)- 1)/((1 - Cos(x))^2) = - ((1 - Cos(x)))/((1 - Cos(x))^2) = (-1)/(1 - Cos(x))`