# Derivative Rules..

• Mar 22nd 2010, 02:21 PM
KnightsDelight08
Derivative Rules..
First all let me say thanks to everyone who helped me last time I came around, I ended up getting an 89% on my first calc test, with most of your help. I'm back again with a few questions, maybe more as im only 50% done my total studying..
but one thing at a time.

I missed the lesson on chain rules, and the book talk is gibberish to me, I learn better with practice . The practice problem I have says:

Find dy/du , du/dx and dy/dx

y=u^2, u =4x+7

I tried working backwards but I don't know where to start, or anything.. any help is appreciated.
• Mar 22nd 2010, 02:32 PM
pickslides
Quote:

Originally Posted by KnightsDelight08

Find dy/du , du/dx and dy/dx

y=u^2, u =4x+7

I tried working backwards but I don't know where to start, or anything.. any help is appreciated.

All you need to use here is simple differentiation

i.e if $y = x^n \implies \frac{dy}{dx} = nx^{n-1}$

$y=u^2 \implies \frac{dy}{du} = 2u$

and

$u =4x+7\implies \frac{du}{dx} = 4$

now using the chain rule

$\frac{dy}{dx}= \frac{dy}{du} \times \frac{du}{dx}$

$\frac{dy}{dx}= 2u \times 4$

$\frac{dy}{dx}= 8u$

$\frac{dy}{dx}= 8 (4x+7)$
• Mar 22nd 2010, 02:35 PM
TKHunny
Your clue is that y = f(u) and u = g(x), this makes y a function of x, but somewhat indirectly. y = f(g(x))

To find dy/dx you must go about it somewhat indirectly.

dy/dx = (dy/du)*(du/dx) That is all. You can sort of see it like du being cancelled from numerator and denominator, but PLEASE don't actually think that way. It's just a short cut to maybe make the formulation easier to recall.

In the other notation. y' = f'(g(x))*g'(x) <== Challenge exploration. WRITE THIS OUT EVERY TIME. You will learn it better.

wrt u, f'(u) is easy. f'(u) = 2u
wrt x, g'(x) is easy. g'(x) = 4

Then dy/dx = f'(g(x))*g'(x) = 2(g(x))*4 = 2(4x+7)*4 = 8(4x+7) = 32x + 56

In this case, it's easy enough to check. Expand f(4x+7) and find its derivative wrt x