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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: kenhung123 on June 10, 2010, 10:01:07 pm

Title: Differentiating inside
Post by: kenhung123 on June 10, 2010, 10:01:07 pm: When we use product, quotient and chain, I find it really confusing to find the derivative of a function inside a function. E.g. $y=tan(sin(x^{2}+4)))$ So using chain rule its quite easy to just differentiate tan and leave the inside but then the next step requires me to multiply by the derivative of the inside using the chain rule. I think its confusing because I need to apply the rule again before I even finished differentiating the first part!
Title: Re: Differentiating inside
Post by: Yitzi_K on June 10, 2010, 10:05:56 pm: That's a chain rule inside a chain rule. Just take it step by step.

It's really $tan(u)$ , where $u=sin(w)$ , and $w=x^2+4$ .

There's nothing difficult there at all. First find $\frac{du}{dx}$ using the chain rule, then it's a simple step to find the derivative of the whole thing.

For something like this it's always easiest to do the chain rule inside-out, rather than outside in.
Title: Re: Differentiating inside
Post by: TrueTears on June 10, 2010, 10:16:48 pm: just do it rigorously

dy/dx = dw/dx * du/dw * dy/du

this reminds me partial derivatives because you often get chains of functions inside each other and u have to keep using the chain rule to partial differentiate them all!
Title: Re: Differentiating inside
Post by: kenhung123 on June 10, 2010, 10:18:03 pm: Oh ok so just define u=inside function? When showing working do I need to say "let u=.."?
Title: Re: Differentiating inside
Post by: Yitzi_K on June 10, 2010, 10:19:26 pm: Quote from: kenhung123 on June 10, 2010, 10:18:03 pm
Oh ok so just define u=inside function? When showing working do I need to say "let u=.."?

yes.
Title: Re: Differentiating inside
Post by: kenhung123 on June 10, 2010, 10:21:14 pm: Quote from: TrueTears on June 10, 2010, 10:16:48 pm
just do it rigorously

dy/dx = dw/dx * du/dw * dy/du

this reminds me partial derivatives because you often get chains of functions inside each other and u have to keep using the chain rule to partial differentiate them all!
I have no idea whats the 3 product of derivatives means. Could you please explain? I understand dy/du*du/dx is when you have function inside function and so you define the inside as u=... and change y=same function but replace u then differentiate both and multiply them so how would I go about the triple derivative
Title: Re: Differentiating inside
Post by: TrueTears on June 10, 2010, 10:38:40 pm: Quote from: Yitzi_K on June 10, 2010, 10:05:56 pm
It's really $tan(u)$ , where $u=sin(w)$ , and $w=x^2+4$ .
it's just chain rule again like yitzi said, the chain rule is not just a technique for 2 functions, it can be used for infinite functions something like y=sin(sin(sin(sin(...)))

so u just keep multiplying until you can get dy/dx

an exercise for you: can u find the general version of the chain rule? if so u can try apply it to this q: write out the chain rule for the case where some function w = f(x,y,z,t) and x=x(u,v) y = y(u,v) z = z(u,v) and t=t(u,v). basically this would be the chain rule for a 4-case scenario!

[the one u learn in methods is a specific 2-case scenario dy/dx = dy/du * du/dx]
Title: Re: Differentiating inside
Post by: kenhung123 on June 10, 2010, 11:25:18 pm: Could you treat the chain rule as the product of the derivatives for every function from outside to inside?
Title: Re: Differentiating inside
Post by: TyErd on June 11, 2010, 12:40:57 am: Is the answer to the problem: $sec^2 sin(x^2 + 4) \times 2x cos(x^2 +4)$ ?
Title: Re: Differentiating inside
Post by: kenhung123 on June 11, 2010, 08:58:44 pm: Yep