How do you differentiate #f(x)= (1 - sin^2x)/(1 - sinx)^2 # using the quotient rule?

Question

How do you differentiate #f(x)= (1 - sin^2x)/(1 - sinx)^2  # using the quotient rule?

Christopher Agresta · Accepted Answer

#(2cosx)/(1-sinx)^2#

Factor the numerator first.

#(1-sinx)(1+sinx)#

Next, remove a factor starting at the bottom.

#((1-sinx)(1+sinx))/((1-sinx)(1-sinx))# #(1+sinx)/(1-sinx)#

We know the quotient rule: #(f'(x)g(x)-g'(x)f(x))/g(x)^2#

So let's insert: #(cosx*(1-sinx)-(-cosx)*(1+sinx))/((1-sinx)^2)#

We get in the numerator: #(cosx*(1-sinx)+cosx*(1+sinx))#

#(cosx(1-sinx+1+sinx))#

#(cosx(2))#

As of right now, overall:

#(2cosx)/(1-sinx)^2#

I double checked this answer, and it can be left this way. There is no more useful algebra to do.

Isaiah Allen · Answer

undefined To differentiate the function $ f(x) = \frac{{1 - \sin^2(x)}}{{(1 - \sin(x))^2}} $ using the quotient rule:

1. Identify $ u(x) $ as the numerator and $ v(x) $ as the denominator.
2. Find the derivatives of $ u(x) $ and $ v(x) $.
3. Apply the quotient rule: $$ \frac{{d}}{{dx}}\left(\frac{{u(x)}}{{v(x)}}\right) = \frac{{u'(x)v(x) - u(x)v'(x)}}{{(v(x))^2}} $$
4. Substitute the derivatives into the quotient rule formula.
5. Simplify the expression.

The derivatives needed are:
- $ u'(x) = \frac{{d}}{{dx}}(1 - \sin^2(x)) $
- $ v'(x) = \frac{{d}}{{dx}}((1 - \sin(x))^2) $

After finding $ u'(x) $ and $ v'(x) $, substitute them into the quotient rule formula and simplify to get the final result.