# Combining differientation rules.. (chain, product, quotient)

• Mar 6th 2010, 11:59 AM
kmjt
Combining differientation rules.. (chain, product, quotient)
I know all 3 of the rules (product, chain, quotient) and I know how to apply them to derive a function when that function ONLY uses 1 of those rules. But questions like this confuse me:

y = [ (2x-1)^2 ] / [ (x-2)^3 ]

Sorry if my brackets look a bit weird i'm trying to make the question clear.

So when I look at that it looks like a quotient rule question AND a chain rule question since the function is to an exponent. How do I apply the rules when I have to use more than one of the rule?
• Mar 6th 2010, 12:16 PM
skeeter
Quote:

Originally Posted by kmjt
I know all 3 of the rules (product, chain, quotient) and I know how to apply them to derive a function when that function ONLY uses 1 of those rules. But questions like this confuse me:

y = [ (2x-1)^2 ] / [ (x-2)^3 ]

Sorry if my brackets look a bit weird i'm trying to make the question clear.

So when I look at that it looks like a quotient rule question AND a chain rule question since the function is to an exponent. How do I apply the rules when I have to use more than one of the rule?

$\displaystyle y = \frac{(2x-1)^2}{(x-2)^3}$

quotient rule, using the chain rule when necessary ...

$\displaystyle \frac{dy}{dx} = \frac{(x-2)^3 \cdot \textcolor{red}{2(2x-1) \cdot 2} - (2x-1)^2 \cdot 3(x-2)^2}{(x-2)^6}$

factor ...

$\displaystyle \frac{dy}{dx} = \frac{(x-2)^2(2x-1)[(x-2) \cdot 4 - (2x-1) \cdot 3]}{(x-2)^6}$

distribute ...

$\displaystyle \frac{dy}{dx} = \frac{(2x-1)[4x-8-6x+3]}{(x-2)^4}$

combine like terms and simplify ...

$\displaystyle \frac{dy}{dx} = \frac{(1-2x)(2x+5)}{(x-2)^4}$
• Mar 6th 2010, 12:20 PM
mathemagister
Quote:

Originally Posted by kmjt
I know all 3 of the rules (product, chain, quotient) and I know how to apply them to derive a function when that function ONLY uses 1 of those rules. But questions like this confuse me:

y = [ (2x-1)^2 ] / [ (x-2)^3 ]

Sorry if my brackets look a bit weird i'm trying to make the question clear.

So when I look at that it looks like a quotient rule question AND a chain rule question since the function is to an exponent. How do I apply the rules when I have to use more than one of the rule?

Since the outside function is a quotient, use the quotient rule while keeping the chain rule in mind:

$\displaystyle \frac{(2x-1)^2}{(x-2)^3}$

Derivative using the quotient rule:

$\displaystyle \frac{(x-2)^3\frac{d}{dx}(2x-1)^2 - (2x-1)^2\frac{d}{dx}(x-2)^3}{((x-2)^3)^2}$

Now use the chain rule for the numerator:

$\displaystyle \frac{(x-2)^32(2x-1)\cdot2 - (2x-1)^2 3(x-2)^2\cdot1}{((x-2)^3)^2}$

And use algebra to tidy up the answer:
$\displaystyle \frac{4(x-2)^3(2x-1) - 3(2x-1)^2 (x-2)^2}{(x-2)^6}$

$\displaystyle \frac{(x-2)^2(2x-1)[4(x-2)-3(2x-1)]}{(x-2)^6}$

$\displaystyle \frac{(2x-1)[4(x-2)-3(2x-1)]}{(x-2)^4}$

$\displaystyle \frac{(2x-1)[-2x-5]}{(x-2)^4}$

$\displaystyle -\frac{(2x-1)(2x+5)}{(x-2)^4}$

$\displaystyle \frac{(1-2x)(2x+5)}{(x-2)^4}$

And there you go! :)
• Mar 6th 2010, 01:21 PM
kmjt
$\displaystyle \frac{dy}{dx} = \frac{(x-2)^3 \cdot \textcolor{red}{2(2x-1) \cdot 2} - (2x-1)^2 \cdot 3(x-2)^2}{(x-2)^6}$

What is confusing me is how do you know what part to use the chain rule on? Other parts in that step are to an exponent, but you are only using the chain rule on the red part why?
• Mar 6th 2010, 01:25 PM
skeeter
Quote:

Originally Posted by kmjt
$\displaystyle \frac{dy}{dx} = \frac{(x-2)^3 \cdot \textcolor{red}{2(2x-1) \cdot 2} - (2x-1)^2 \cdot 3(x-2)^2}{(x-2)^6}$

What is confusing me is how do you know what part to use the chain rule on? Other parts in that step are to an exponent, but you are only using the chain rule on the red part why?

because it's the only part that really needed it ... the derivative of $\displaystyle (2x-1)^2$

the derivative of $\displaystyle (x-2)^3$ is $\displaystyle 3(x-2)^2 \cdot 1$
• Mar 6th 2010, 02:12 PM
kmjt
How do I know what needs the chain rule and what doesn't? Sorry.
• Mar 6th 2010, 03:14 PM
skeeter
Quote:

Originally Posted by kmjt
How do I know what needs the chain rule and what doesn't? Sorry.

The Chain Rule
• Mar 6th 2010, 03:48 PM
db5vry
Quote:

Originally Posted by kmjt
How do I know what needs the chain rule and what doesn't? Sorry.

Providing you know/have memorised the formulas for each of the differentiation rules, and are able to recognise when to apply them, then it's not so bad.

Lets, say, for example, the product rule is u(dv/dx) + v(du/dx), where you have two seperate functions, u and v. If v, for example, was something like $\displaystyle (x-4)^{5}$ then what do you need to do to differentiate it? You need the chain rule. Using the chain rule gives $\displaystyle 5(x-4)^4$ and if you don't know how this works, you could wisely spend some time going back over the chain rule.

You would need this to put back into the product rule formula for the part of (dv/dx) - so here, you can see that one part of the product rule example here includes an application of the chain rule. As examples get more complicated, the more you might have to recognise this.

So in summary, it will help you to write out what u and v actually are, consider them both, and if you need the differential of one of these to put into another formula and need to consider one of the other rules of differentiation to what you are primarily using, then it helps to consider each function on its own first before you proceed any further towards solving problems. I hope I helped :)