Find the derivative

[CrazCo]

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.

[DJBruce]

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.

[CrazCo]

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)

[mooeypoo]

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.

[DJBruce]

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, 2009 by DJBruce
corrected Tex.

[mooeypoo]

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

[CrazCo]

Thank you, I got it!