Question

find the derivative of f(x)=(cosx/1-sinx)^2
steps to finding this answer?

Answer 1

Apply the chain rule followed by the quotient rule.

Answer 2

Where are you getting stuck?

Answer 3

On the chain rule part

Answer 4

Okay. The chain rule is \((f\circ g)'(x)=f'(g(x))\cdot g'(x) \), right? So, we need to figure out what \(f\) and \(g\) are in this problem. If \(f(x)=\left(\dfrac{\cos x}{1-\sin x}\right)^2\), we can take \(f(x)=x^2\) and \(g(x)=\dfrac{\cos x}{1-\sin x}\). Do you see how that part works?

Answer 5

If \(f(x)=x^2\) and \(g(x)=\dfrac{\cos x}{1-\sin x}\), then \(f(g(x))=\left(\dfrac{\cos x}{1-\sin x}\right)^2\), our original formula.

Answer 6

@dilfalicious Are you with me so far?

Answer 7

first simplify f(x) so:
f(x)=cos^2x/(1-sinx)^2
now recall cos^2x=1-sin^2x=(1-sinx)(1+sinx)
f(x)=(1-sinx)(1+sinx)/(1-sinx)(1-sinx)=(1+sinx)/(1-sinx)
now apply only quotient rule so
f'(x)=((1-sinx)D(1+sinx)-(1+sinx)D(1-sinx))/(1-sinx)^2
=(1-sinx)(cosx)-(1+sinx)(-cosx)/(1-sinx)^2
=(cosx-sinxcosx+cosx+sinxcosx)/(1-sinx)^2
=2cosx/(1-sinx)^2 <answer

Answer 8

The way @anonymoustwo44 did it will work, but you should also learn how to use the chain rule, because you'll need to use it often.

Answer 9

what would be the final step?

Answer 10

Which step is the final one depends on which order you do things in. What do you have so far?

Answer 11

I'm not sure where to go from your last step you posted

Answer 12

Well, you know now what \(f(x)\) and \(g(x)\) are, so you apply the Chain Rule formula. Find \(f'(g(x))\cdot g'(x) \)

Answer 13

so the derivative of f times g(x)?

Answer 14

No, the derivative of f of g of x, times the derivative of g of x. So you take the derivative of f(x) then plug in g(x), and multiply that by the derivative of g(x)

Answer 15

I have to go to work, I'll be back on in about 8 hrs or so.

Answer 16

\[(\cos2x/1-\sin2)\times g \prime(x) ?\]

Answer 17

thanks for your help

Answer 18

Remember f(x) is just x^2, so f'(x) is just 2x, then substitute g(x) in for x, so 2g(x). Then mutiply by g'(x) Good luck, I'll be back on later