# Using the Chain Rule

1. Oct 15, 2006

Using the "Chain Rule"

Alright I've been working on some more practice problems and doing fairly well in finding the correct derivative, until I came to this question:

y = cos(sin^2x) (everything is under the square root sign)

We know that f'(x) = 1/2X^-(1/2) or 1/(2(root)X) and g'(x) = cos(sin^2x) I believe.

But I'm having troubles trying to imply our chain rule into the problem. Can anyone help me try and interpret this derviative? I just want to say that you guys have been great in helping me understand my mistakes.

2. Oct 15, 2006

$$y = \sqrt{\cos(\sin^{2}x)}$$. Let $$u = \cos(\sin^{2} x)$$. Then $$y = \sqrt{u}$$

Then $$\frac{dy}{dx} = \frac{1}{2}u^{-\frac{1}{2}} \frac{du}{dx}$$. What is $$\frac{du}{dx}$$?

Last edited: Oct 15, 2006
3. Oct 15, 2006

du/dx would be -sin(sin^2x)*2*sinxcosx right?

* meaning multiply

4. Oct 15, 2006

that is correct.

5. Oct 15, 2006

Alright sweet, so now we have to use the chain rule I believe? However, we have to use our du's right, and I'm not sure what steps to do from here.

Last edited: Oct 15, 2006
6. Oct 15, 2006

$$\frac{dy}{dx} = \frac{1}{2}u^{-\frac{1}{2}} \frac{du}{dx}$$

This is the chain rule. multiply $$\frac{du}{dx}$$ by $$\frac{1}{2}u^{-\frac{1}{2}}$$.

We really have $$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}$$

7. Oct 15, 2006

Ohhh okay I see what you are saying now. Gotcha amigo! Thanks!

8. Oct 15, 2006

One last little question I've got for you for now, thanks a lot everyone, especially you courtrigrad, you've helped out a lot. It asks y= cos(pi)x/ (sin(pi)x+cos(pi)x)

We know f(x) = cos(pi)x and g(x) = sin(pi)x+cos(pi)x

y' = dy/dx = g(x)f'(x)-f(x)g'(x)/ g(x)^2

And we know that the f'(x) = -1 since cos(pi)x = -1x and g'(x) = -1 since sin(pi)x+cos(pi)x= 0x+-1x = -1x

Subbing into our formula, we get:
sin(pi)x+cos(pi)x * (-1) - cos(pi)x * (-1)/ (sin(pi)x+cos(pi)x)^2 I believe, however, what's the next step from here in solving this derivative.

9. Oct 15, 2006

$$y = \frac{\cos(\pi x)}{\sin \pi x + \cos \pi x}$$.

$$f'(x)$$ does not equal 1. It is equaled to $$f'(x) = -\pi\sin \pi x$$.

and $$g'(x) = \pi\cos \pi x - \pi\sin \pi x$$

10. Oct 15, 2006

Ahhh okay and then you sub those into our formula to get:
(sin(pi)x+cos(pi)x)(-pi(sin(pi)x) - cos(pi)x*(pi(cos(pi)x-pi(sin(pi)x) / (sin(pi)x+cos(pi)x)^2

Well basically what you said, but subbing them into our formula, it's hard to type it all out haha. But simplifying this expression proves difficult for me, you basically multiply to get everything out of the brackets, and I've even tried conversing with a few friends but they've been no help in answering, so even if you just show the next step after subbing in all our f(x), g(x), f'(x) and g'(x) values for simplifying this expression, I'm sure I can figure out the rest. You've been a great help though and real patient with me so thank you, if I could I'd give you a cookie haha.

Last edited: Oct 15, 2006
11. Oct 15, 2006

$$\frac{(\sin \pi x + \cos \pi x)(-\pi\sin \pi x) - \cos \pi x(\pi\cos \pi x-\pi \sin \pi x)}{(\sin \pi x +\cos \pi x)^{2}}$$

Expand it out:

$$\frac{-\pi \sin^{2} \pi x - \pi \sin \pi x\cos \pi x - \pi \cos^{2} \pi x + \pi \sin \pi x \cos \pi x}{\sin^{2} \pi x + 2\sin \pi x\cos \pi x + \cos^{2} \pi x}$$.

and use the identity $$\sin^{2} x + \cos^{2} x = 1$$

you should get: $$\frac{-\pi}{2\sin \pi x \cos \pi x + 1}$$

Last edited: Oct 16, 2006
12. Oct 16, 2006

Alrighty here's what I ended up with from that, since the (PI)sin(PI)xcos(PI)x's cancel out, we are left with:
= -(PI)[sin^2x + cos^2x] / sin^2(PI)x+2sin(PI)xcos(PI)x+cos^2(PI)x
And using our identity sin^2x + cos^2x = 1, we get,
-(PI) / sin^2(PI)x+2sin(PI)xcos(PI)x+cos^2(PI)x
= -(PI) / (sin^2(PI)x + cos^2(PI)x) + 2sin(PI)xcos(PI)x

13. Oct 16, 2006

the $$\sin^{2}\pi x + \cos^{2} \pi x = 1$$ in the denominator. left out the 1, sorry.

Last edited: Oct 16, 2006
14. Oct 16, 2006