Quick Differentiation Question (Trig).

Am I correct in thinking the quotient rule must be used?

no, you can write it as $cosx(1-sinx)^{-1}$ and use the product and chain rules

Oh ok, that's fine, thanks

Could you please tell me why you can't use the quotient rule? Is it the presence of the 1-sinx?

i didn't say you can't use the quotient rule, either method will work

You can use it but what he suggested is much easier and it would be prudent to be able to spot this!

Wait, having used that i'm still having trouble.

cosx(1-sinx)^-1 I differentiate to -cosx-cosx(1-sinx)^-2 -> cos^2x/1-sin^2x

I think i'm going wrong somewhere but i'm not sure where?

Use chain rule on (1-sin(x))^-1 and then product rule on cos(x)*(1-sin(x))^-1. Post further working if you get stuck

Argh i'm getting all mixed up in (1-sinx)^-3 which I know can't be right!

So I have cosx(1-sinx)^-1

Using the chain rules appears to give me -1 x cosx x (1-sinx)^-2 x -cosx which eventually cancels to 1, which obviously isn't right.

Where am I going wrong here, have I missed a step?

(1-sin(x))^-1 differentiated gives cos(x)*(1-sin(x))^-2

this is your du/dx value.

V = cos(x)

v*du/dx = -sin(x)/1-sin(x)

u*dv/dx = cos^(2)x/(1-sin^(2)x)

I was stating U = cos(x), don't know if that makes a difference?

Anyway, so from that we deuce that Vdu/dx + Udv/dx =>

-sinx/1-sinx + 1 but that doesn't come to 1/1-sinx I don't think?

y = cosx/1-sinx , find dy/dx

I know the answer is 1/1-sinx but I just can't get to it. Am I correct in thinking the quotient rule must be used?

Thanks in advance.