of y=cos^2[(1-sqrtx) / (1+sqrtx)]?

 

I tried using the power rule, then trig derivative of cos and quotient rule for the argument but i got it wrong.

That's a nasty looking derivative, however you can take it. You will need to use the chain rule and the quotient rule. If you post your work I will try and see where you went wrong.

y=cos^2[(1-sqrtx) / (1+sqrtx)]?

 

2[cos(1-sqrtx) / (1+sqrtx)] * -sin[(1-sqrtx) / (1+sqrtx)] * [(1+sqrtx)(-1/2x^-1/2)-(1-sqrtx)(1/2x^-1/2)] / (1+sqrtx)^2)

Think of it this way:

 

[math]y=cos^2[\frac{v(x)}{u(x)}][/math]

 

Where

 

[math]v(x)=1-\sqrt{x}[/math]

[math]u(x)=1+\sqrt{x}[/math]

 

And so, now doing the chain-rule should be easier. Here's the start (continue on your own):

 

[math]\frac{dy}{dx}=-2cos(\frac{v(x)}{u(x)})sin(\frac{v(x)}{u(x)})\big( \frac{\frac{dv}{dx}u(x) + \frac{du}{dx}v(x)}{u(x)^2} \big)[/math]

 

And so now you need to figure out what du/dx and dv/dx are (the derivatives of the parts of the fraction) and fill in the gaps, more or less.

 

Now, I believe I did it right, but it's 1:20am here, so I might've had some mistakes.. make sure you go over everything. The point of my suggestion is more for order's sake -- when you make your exercise organized, it's much easier to solve and much less scary.

 

by the way, you can use "styled" math by using [math]y=[/math] tags.

That is almost what I got to Mooey, however, I think it should be:

 

[math]

\frac{dy}{dx}=-2cos(\frac{v(x)}{u(x)})sin(\frac{v(x)}{u(x)})\big( \frac{\frac{dv}{dx}u(x) -\frac{du}{dx}v(x)}{u(x)^2} \big)

[/math]

 

I think that is what he got, however it is somewhat hard to tell. If your answer still isn't write they might be giving a simplified answers. So you might want to remove any complex fractions and simplify any trigonometric identities, hint hint double angle identities.

Edited November 28, 200916 yr by DJBruce
corrected Tex.

Whoops, You're right, sorry, there should be a minus on top, not a plus. Sorry 'bout that.

Thank you, I got it!