Find derivative of y = [ (3x - 2)^{-2} ] / [ (2x - 6)^{-2} ] using Quotient Rule, Chain Rule

[ADG420]

Could someone help me find this derivative using (1) the Quotient Rule and (2) the Chain Rule?

 

[stapel]

Could someone help me find this derivative using (1) the Quotient Rule and (2) the Chain Rule?

[imath]\qquad y = \dfrac{(3x - 2)^{-2}}{(2x - 6)^{-2}}[/imath]

We'll be glad to help! But first we'll need to see what you've tried and how far you've gotten.

Thank you!

 

[Steven G]

Hint: [(3x-2)^-2]]'=-2(3x-2)^-3*3

You do know that math is not a spectator sport. Now show us your work so the tutors here can help.

 

[JeffM]

First, the problem formulation itself gives you a hint. You are to use the chain rule. That suggests a substirution. Of course it does not sugesst which substitution. Selecting USEFUL substitutions is an art that you develop with experience.

Second, I shall give a hint on a helpful substitution.

[math]\text {Let } p = (3x - 2)^2 \text { and } q = (2x - 6)^2.[/math]

 

[ADG420]

This is where I left off. I can't proceed becaus I don't know how to multiply v*u' and u*v'.
I'm also worried about v^2 because there are two exponents.

 

[ADG420]

What do I do with ((2x-6)^-2)^2?

 

[ADG420]

 

[Dr.Peterson]

What do I do with ((2x-6)^-2)^2?

Apply the fact that [imath](a^m)^n=a^{mn}[/imath].

This is where I left off. I can't proceed because I don't know how to multiply v*u' and u*v'.

You're referring to this:

​

First, you need to correct it, because these are multiplications, not subtractions:

​

Then I would factor out the lowest power of each base, namely [imath](2x-6)^{-3}(2x-2)^{-3}[/imath]. Or, if you prefer, multiply the numerator and denominator by [imath](2x-6)^{3}(2x-2)^{3}[/imath] to eliminate the negative powers.

I often observe that calculus is where you finally learn algebra, because you have to use it as a tool. All of this is things you are expected to know already, but weren't forced to really master.

 

[JeffM]

First, the problem formulation itself gives you a hint. You are to use the chain rule. That suggests a substirution. Of course it does not sugesst which substitution. Selecting USEFUL substitutions is an art that you develop with experience.

Second, I shall give a hint on a helpful substitution.

[math]\text {Let } p = (3x - 2)^2 \text { and } q = (2x - 6)^2.[/math]

You clearly read my post #4 because you thanked me for it, and you're welcome.

But you did not use it.

Substitution is NEVER necessary in differential calculus, but it is frequently very useful.

[math]\text {Let } p = (3x - 2)^2 \text { and } q = (2x - 6)^2 \implies\\ p'= 2(3x - 2)(3) = 6p \text { and } q' = 2(2x - 6)(2) = 4q.\\ \text {Given: } y = \dfrac{(3x - 2)^{-2}}{(2x - 6)^{-2}} = \dfrac{p^{-1}}{q^{-1}} = \dfrac{q}{p}.\\ \therefore \ y' = \dfrac{qp' - q'p}{p^2} = \dfrac{q(6p) - 4qp}{p^2} = \dfrac{6q - 4q}{p} = \dfrac{2q}{p} = \text {WHAT?}[/math]
Now I completely agree with Dr. Peterson that many people learn algebra by studying calculus. I did for one. But the substitution I suggested is one way to make the algebra far less error prone. It is a bit more time consuming, but you will make fewer errors.

 

[Steven G]

What do I do with ((2x-6)^-2)^2?

If you don't know/remember, then you go back to the definitions...

Consider (x²)³= (x²)(x²)(x²)=(xx)(xx)(xx) = xxxxxx=x⁶

Now can you see what you could do with the two initial powers to get 6? Look at the blue part and think.

 

[Steven G]

I thought that I knew my algebra until I took calculus.

I would have used algebra before taking the derivative.

\(\displaystyle \dfrac{(3x-2)^{-2}}{(2x-6)^{-2}} =( \dfrac{(3x-2)}{(2x-6)})^{-2}\ OR\ ( \dfrac{(2x-6)}{(3x-2)})^{2}\ =\ 4( \dfrac {(x-3)}{(3x-2)})^{2}\)

Now take the derivative.

 

[Steven G]

I think that the product and quotient rule should be written as I have below, so that they look similar!

\(\displaystyle (uv)' = u'v + uv'\)

\(\displaystyle (u/v)'\ =\dfrac{ u'v - uv'}{v^2}\)

Now the numerators are the same except for the middle sign. In my opinion, this is much easier to memorize.

 

[khansaheb]

I thought that I knew my algebra until I took calculus.

I would have used algebra before taking the derivative.

\(\displaystyle \dfrac{(3x-2)^{-2}}{(2x-6)^{-2}} =( \dfrac{(3x-2)}{(2x-6)})^{-2}\ OR\ ( \dfrac{(2x-6)}{(3x-2)})^{2}\ =\ 4[ \dfrac {(x-3)}{(3x-2)}]^{2}\)

Now take the derivative.

I would have gone one step further:

\(\displaystyle \dfrac{(3x-2)^{-2}}{(2x-6)^{-2}} =( \dfrac{(3x-2)}{(2x-6)})^{-2}\ OR\ ( \dfrac{(2x-6)}{(3x-2)})^{2}\ =\ 4( \dfrac {(x-3)}{(3x-2)})^{2} = \ \frac{4}{9}[ \dfrac {(3x-9)}{(3x-2)}]^{2} = \ \frac{4}{9}[ 1-\dfrac {7}{(3x-2)}]^{2} \)

I know it looks a bit more horrendous (than response #11), but I "believe" differentiation becomes a bit easier - not by much though!

 

[khansaheb]

I think that the product and quotient rule should be written as I have below, so that they look similar!

\(\displaystyle (uv)' = u'v + uv'\)

\(\displaystyle (u/v)'\ =\dfrac{ u'v - uv'}{v^2}\)

Now the numerators are the same except for the middle sign. In my opinion, this is much easier to memorize.

That is how I have in "my head". I re-derive (u/v)' in my head every-time I need it, because I keep forgetting the "negative" sign in the numerator.