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I am reading a book called "Elementary Calculus: An Approach Using Infinitesimals" in my spare time. On page 69 there is the problem - y=(x^2+5)^3 How do I solve it? I get the answer 3(x^2+5)^2 yet my calculater says it is 6x(x^2+5)^2. My calculater is correct (I graphed the line and my dirivitive and the calcs dirivitve and the calculater gives a much better answer. My question is, how did it get that answer? Thank you for your time, Arrummzen

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Use the chain rule.

Basically, if you have a function that is complex, such as (x^2+5)^3 you can''t simply do the shortcut once since you haven''t actually differentiated the x term fully.

Essentially, you need to think of that function as y = u^3 (where u = (x^2+5)) and then take the derivative of u with respect to x (written as du/dx).

This way, to find dy/dx you need to find dy/du*du/dx.

So, you get 3*(2x)*(x^2+5)^2 which is the correct answer your calculator gave you.

I hope this helps.

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You need to take the derv of what's inside the () as well:

y = (x^2+5)^3
y' = 3(x^2+5)^2 * dy/dx(x^2+5)
y' = 3(x^2+5)^2 * 2x
y' = 6x(x^2+5)^2

=============
Andrew

[edited by - acraig on June 9, 2004 7:06:59 PM]

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If you act like that''s two functions, instead of one, it''s easier. Think of it as y = f(g(x)); where f(x) = x3 and g(x) = x2 + 5;

From there just use the chain rule. Using that I get 6x(x2 + 5)2

And for future reference, the mods don''t like homework questions here. But there are some sites in the forum FAQ that can help you out next time.

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Just had my pure maths 3 exam 2day (i think ive failed) which had all this really hard differentation and intergration, its been doing my head in lol.

He said he was reading it in his spare time, not for homework.

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Now why exactly does this chain rule work? I can''t seem to find any equivilent in my book...

Yea, this isn''t for homework. im trying to learn how do program computer games, and I have some game programing books. Most of the math I understand, but some of the stuff (Id say about 10% of it) I need calculus to understand, so I am hopeing to pick up a little calculus over the summer. I just finished algebra2 and next year I am taking trig/precalc.

Arrummzen

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The chain rule works like this:

You have a function, say... (x2 + 2x)3, and as you can see, that function can be composed of two smaller functions, f(x) and g(x) where g(x) = x2 + 2x and f(x) = x3.

Then, when you write the function using the two smaller functions you get f(g(x)). Normally when you find the derivative of a function, you find the change in y over the change is x (dy/dx), but this time you''re finding the change in y over the change in g, but you still need to find the change in g over the change in x. So in the end you have to find both, and the chain rule comes out to be this: f''(g(x))g''(x), or dy/du * du/dx in Leibniz notation. (du is basically the g(x), so it''s the change in y with respect to the change in g(x) times the change in g(x) with respect to the change in x)

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Heh good old P3, so very very simple. Actually to be fair I haven''t had to differentiate anything harder than in P3 in my first year of maths at uni. But then applied maths is more applicable to a physics course.

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quote:
Original post by Arrummzen
I am reading a book called "Elementary Calculus: An Approach Using Infinitesimals" in my spare time. On page 69 there is the problem -
y=(x^2+5)^3
How do I solve it? I get the answer 3(x^2+5)^2 yet my calculater says it is 6x(x^2+5)^2. My calculater is correct (I graphed the line and my dirivitive and the calcs dirivitve and the calculater gives a much better answer. My question is, how did it get that answer?

Arrummzen

That must be a pretty fancy calculator you''ve got there if it can calculate derivates for you using symbols.

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Were can I find a proof of the chain rule (a good simple proof, nothing to complicated)?

Arrummzen

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you have df(g(x))/dx, by definition, is:

lim (f(g(x))-f(g(x0)))/(x-x0)
x->x0

if you multiply the numerator and the denominator by
(g(x)-g(x0), you get:

lim ((f(g(x))-f(g(x0)))*(g(x)-g(x0)))/((x-x0)*(g(x)-g(x0))
x->x0

just ordering the things different, you get:
lim     f(g(x))-f(g(x0))*g(x)-g(x0)=df/dg*dg/dxx->x0       g(x)-g(x0)     x-x0

Now, it is easy to see that:
df/dx=df/dg*dg/dx

[edited by - danielf on June 9, 2004 8:42:13 PM]

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quote:
Original post by SpaceDude
That must be a pretty fancy calculator you''ve got there if it can calculate derivates for you using symbols.

Probably just a TI-89.

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danielf''s "proof" is not valid. g(x0)-g(x) could be 0 in many, many points, which would make the whole argument fail.

You can find a better proof here, for instance: http://www.shu.edu/projects/reals/cont/proofs/diffalg.html