How do you differentiate f(x) = (sinx)/(1-cosx) using the quotient rule?

1 Answer
Mar 17, 2018

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

Explanation:

The Quotient Rule for derivatives states:

If f(x) = g(x)/(h(x))

then

f'(x) = (g'(x)h(x) - g(x)h'(x))/(h(x))^2

SEE: https://en.wikipedia.org/wiki/Quotient_rule

So, the first step is to identify g(x) and h(x) in your original function:

f(x) = sin(x)/(1- cos(x))

so

g(x) = sin(x)

h(x) = 1- cos(x)

We then differentiate the two components of f(x) with respect to x:

g'(x) = cos(x)

h'(x) = -(-sin(x)) = sin(x)

Let's combine all of the pieces, based on the Quotient Rule, and then simplify:

f'(x) = (g'(x)h(x) - g(x)h'(x))/(h(x))^2

f'(x) = (cos(x)(1- cos(x)) - sin(x) sin(x))/(1- cos(x))^2

f'(x) = (cos(x) - cos^2(x) - sin^2(x))/(1- cos(x))^2

f'(x) = (cos(x) - (cos^2(x) + sin^2(x)))/(1- cos(x))^2

f'(x) = (cos(x) - 1)/(1- cos(x))^2

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

f'(x) = cancel(cos(x) - 1)/(cancel((cos(x) - 1))*(cos(x) - 1)

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