Master the Chain Rule: A Guide to Finding Derivatives of Complex Functions

[loadsy]

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 = [PLAIN]http://upload.wikimedia.org/math/1/6/0/160a5b4ac79375bd4c5e13c0f3a95f73.pngcos(sin^2x)[/URL] (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.

 

[courtrigrad]

$$y = \sqrt{\cos(\sin^{2}x)}$$. 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 $$\frac{du}{dx}$$?

 

[loadsy]

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

* meaning multiply

 

[courtrigrad]

that is correct.

 

[loadsy]

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.

 

[courtrigrad]

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

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

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

 

[loadsy]

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

 

[loadsy]

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.

 

[courtrigrad]

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

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

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

 

[loadsy]

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.

 

[courtrigrad]

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

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

 

[loadsy]

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

 

[courtrigrad]

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

 

[loadsy]

Alright, so it just becomes -(PI)/ 1 + 2sin(PI)xcos(PI)x then, yeah I wasn't too sure whether or not we could use our identity for both the numerator and denominator, alright thanks for your help. :D I just want to say that you've done a fantastic job in helping me answer my problems, even some really silly mistakes you've noticed too along the way with 2/7 rather than 2/3 for that one tangent question, I was going too fast I guess and got carried away, but thanks a bunch.