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

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Answer 1 · 2018-04-23T04:35:34

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

First, factor the numerator.

$(1-sinx)(1+sinx)$

Then cancel a factor from 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:
$(cosx*(1-sinx)+cosx*(1+sinx))$

$(cosx(1-sinx+1+sinx))$

$(cosx(2))$

Now in total:

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

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

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