One more derivative problem

1. Sep 9, 2007

fitz_calc

This is an example I found online. I know how to get to this: , but why is 3-4x brought into the problem? sorry for the newbie questions tonight but my notes from class don't really cover the material very well, thanks.

2. Sep 9, 2007

rocomath

sorry for late response, server was lagging or something

chain rule, let me type this out ... refresh in like 2 min.

$$f(x)=(x-3x^2)^n$$ <-example

bring the power down, subtract it by 1, then do a "chain" of the base and take it's derivative

$$f'(x)=n(x-3x^2)^{n-1}(x-2\times3x^{2-1})$$ <-example

$$f(x)=(3x-2x^2)^3$$

$$f'(x)=3(3x-2x^2)^2(3-4x)$$

$$f'(x)=(9-12x)(3x-2x^2)^2$$

Last edited: Sep 9, 2007
3. Sep 9, 2007

As rocophysics has shown, it is just the chain rule. The way I learned the chain rule was qualitatively (through words) and it helped me out.

It may seem a little "wordy at first", but it can help if you say it out as you actually do it.

If you have a composite of functions, that is an "outside function" and an "inside function"---->in this case, the cubic function is the "outer" and the quadratic in the parenthesis is the inside----->

then the derivative of the composite function is:
(the derivative of the "outside function", with the "inside function" as its argument) times (the derivative of the "inside function")

....and you can keep on applying this for any number of functions within a composite function. Just keep working from the outside to the inside.

Hope that helps,
Casey

4. Sep 9, 2007

rocomath

that's a great way Casey, makes a lot of sense too; exactly how i learned logs

5. Sep 9, 2007

Same here with logs. Sometimes being able to actually say what it is that you are really doing while doing it makes a big difference.

Casey

6. Sep 9, 2007

fitz_calc

ok i think i'm getting it. i tried a homework problem:

y=(4+x^4)^5 => u=g(x)=4+x^4 , f(u)=u^5
then f(x)=f(g(x)) => 5u^4 * (4x^3) => 5(4+x^4)^4 * (4x^3)

* a bit of confusion here -- why can't I multiply the 5 by (4+x^4)^4?

I know the answer is 20x^3 (4+x^4)^4 but can't figure out why the 5 isn't distributed in the last step, thanks

7. Sep 9, 2007

rocomath

b/c it's raised to a power, i don't really know what to say to be more specific, but i'm sure someone else will give you a better answer

though, x^3 is raised to a power as well, it's not the same, lol sorry

8. Sep 9, 2007

fitz_calc

i'm not sure either -- i just did another similar problem and sure enough the same method was applied to that solution as well. i'll wait for a response

9. Sep 9, 2007

rocomath

it's basically algebra

$$a^{m} \times a^{n} = a^{m+n}$$ same base, add their exponents

$$(ab)^{m} \times a^{n}$$ does not equal $$(ab)^{m+n}$$

take this

$$5 \times (a+b)^2 = 5(a+b)^2$$ if you want to multiply through, you'll have to expand

$$5(a^2+2ab+b^2) = 5a^2 + 10ab + 5b^2$$

Last edited: Sep 9, 2007
10. Sep 9, 2007

Feldoh

$$\frac{d}{dx} f(g(x)) = f'(g(x))*g'(x)$$

That's why you get 3-4x

For example $$h(x) = sin(x^2)$$ find the first derivative

We'd call $$f(x) = sin(x)$$ and $$g(x) = x^2$$

Since $$h(x) = f(g(x))$$ we just follow the chain rule.

$$f'(x) = cos(x)$$ and $$g'(x) = 2x$$

So we get

$$2x*cos(x^2)$$

11. Sep 10, 2007

Gib Z

Sometimes the chain rule is written in "Leibniz differential" notation, you may have seen it before.

$$\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx}$$.

In that notation, we have to make a suitable expression for u, and its always the "inside function".

12. Feb 22, 2009

MrSparky

Well basically to find the derivative its (derivative of original equation) . (power of original equation) . ( original equation)^(power of original equation - 1), in this case derivative of original equation is 4x^3 , power of original equation is 5, (original equation^ power of original equation - 1) is (4+4^4)^(5-1)

Now using what i wrote above the derivative is (4x^3) . (5) . (4+x^4)^4

13. Feb 23, 2009