Trouble with chain/product/quotient rule

[Silencher]

Hello,

Here is the problem I'm trying to solve:
[ (x-6)/(x+9)]^4

I'm really not sure what to do. I tried just solving for the inner/fractional part first, then I thought I could apply the chain rule, but... it's just a mess. Here's what I have so far:


Step 1: Bring denominator up


[(x-6)(x+9)^(-1)]^4


Step 2: Set g/h/j


g(x): (x+9)^(-1)
h(x): (x-6)
g'(x): (-1)(1)(x+9)^(-2)
h'(x): 1
j(x): [stuff]^4
h/g': ((x+9)^(-1)) +(-x+6)((x+9)^(-2))


j'(x): 4[(x+9)^(-1) +(-x+6)((x+9)^(-2))]^3 <--answer?

Any help would be appreciated. I'm really terrible at math/calculus especially, and I almost always split up the problem as I did up above. If someone could show me step by step where I'm going right and where I'm going wrong, that would be great. I'd be fine if it was a different problem in the same format, just so I can get the concept down of how to derive a fraction set within an exponent?

 

[Jason76]

Post Edited 9/28/13

I'll give it a shot:

\(\displaystyle f(x) = [\dfrac{(x-6)}{(x+9)}]^4\)

\(\displaystyle f'(x) = 4^3\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{(x + 9)(1) - (x - 6)(1)}{(x + 9)^{2}}]\) - \(\displaystyle u(du)\). For \(\displaystyle du\), use the quotient rule: \(\displaystyle g(f') - f(g')/g^{2}\) Given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{(x + 9) - (x - 6)}{(x + 9)^{2}}]\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{x + 9 - x + 6}{(x + 9)^{2}}]\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{3}{(x + 9)^{2}}]\)

\(\displaystyle f'(x) = 4[\dfrac{(x-6)}{(x+9)}]^{3}[\dfrac{3}{(x + 9)^{2}}]\) - Back Substitute

 

[Silencher]

question

Before reading below, it looks like you're still working on your answer, Jason, so sorry, let me wait for you to finish.


---

Oh I should have been using the quotient rule... right.

But isn't the quotient rule you have off? I have it as:

bottom_prime*top - top_prime*bottom
_________________________
bottom^2

But yours looks like you're using + rather than -?

So I have:

4[(-15)/((x+9)^2)]^3

But I'm not sure if that's correct either.

 

[Jason76]

Before reading below, it looks like you're still working on your answer, Jason, so sorry, let me wait for you to finish.


---

Oh I should have been using the quotient rule... right.

But isn't the quotient rule you have off? I have it as:

bottom_prime*top - top_prime*bottom
_________________________
bottom^2

But yours looks like you're using + rather than -?

So I have:

4[(-15)/((x+9)^2)]^3

But I'm not sure if that's correct either.

The original post has been edited.

 

[Silencher]

Thank you!

I think I follow the logic of how it was reached, I have one last problem (in another thread) and I'll know whether or not this is correct, but I'm sure it is, I re-did the problem and I came to the same answer. I will edit this note once I know.

 

[Jason76]

Actually, I'm not really sure if using the quotient rule is correct here. Perhaps you just take the derivative of the bottom and the top (as du) and then multiply to u. Can an expert on here clear this up? Thanks.

Is it?

\(\displaystyle f(x) = [\dfrac{(x-6)}{(x+9)}]^4\)

\(\displaystyle f'(x) = 4^3\)

\(\displaystyle f'(x) = 4[\dfrac{(x-6)}{(x+9)}]^{3} \dfrac{1}{1}\)

\(\displaystyle f'(x) = 4[\dfrac{(x-6)}{(x+9)}]^{3} (1)\)

\(\displaystyle f'(x) = 4[\dfrac{(x-6)}{(x+9)}]^{3}\)

 

[daon2]

Jason, why is it you're often so careless? In each and every line of mathematics you write, there should be no guessing. Writing an equal sign says "with absolute certainty, these are the same." Make sure everything you write you are certain about. This is for yourself and to save confusion of those you are trying to help. Your first answer is correct, but again, the derivation was done carelessly.

If \(\displaystyle u=(x-6)/(x-9)\), then using the quotient rule \(\displaystyle u' = -3/(x-9)^2\).

Any time a substitution is made, the chain rule applies.

\(\displaystyle f'(x) = f'(u)\cdot u'\)

\(\displaystyle f'(x) = 4u^3\cdot u' = 4\cdot [(x-6)/(x-9)]^3\cdot [-3/(x-9)^2]\)

 

[Jason76]

Jason, why is it you're often so careless? In each and every line of mathematics you write, there should be no guessing. Writing an equal sign says "with absolute certainty, these are the same." Make sure everything you write you are certain about. This is for yourself and to save confusion of those you are trying to help. Your first answer is correct, but again, the derivation was done carelessly.

If \(\displaystyle u=(x-6)/(x-9)\), then using the quotient rule \(\displaystyle u' = -3/(x-9)^2\).

Any time a substitution is made, the chain rule applies.

\(\displaystyle f'(x) = f'(u)\cdot u'\)

\(\displaystyle f'(x) = 4u^3\cdot u' = 4\cdot [(x-6)/(x-9)]^3\cdot [-3/(x-9)^2]\)

In the original post, I did the quotient rule. But you simplified it further. I think we used the quotient rule, because we dealt with two fractional terms in parenthesis. But if there were no parenthesis, then we would NOT use the quotient rule. Is this correct?

\(\displaystyle f(x) = [\dfrac{(x-6)}{(x+9)}]^4\)

\(\displaystyle f'(x) = 4^3\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{(x + 9)(1) - (x - 6)(1)}{(x + 9)^{2}}]\) - \(\displaystyle u(du)\). For \(\displaystyle du\), use the quotient rule: \(\displaystyle g(f') - f(g')/g^{2}\) Given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{(x + 9) - (x - 6)}{(x + 9)^{2}}]\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{x + 9 - x + 6}{(x + 9)^{2}}]\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{3}{(x + 9)^{2}}]\)

\(\displaystyle f'(x) = 4[\dfrac{(x-6)}{(x+9)}]^{3}[\dfrac{3}{(x + 9)^{2}}]\) - Back Substitute

 

[Silencher]

thanks!

Like I said, I tried this on my own and came to the same answer, Jason and when I submitted it was correct. So thank you very much.

 

[lookagain]

In the original post, I did the quotient rule. But you simplified it further.

I think we used the quotient rule, because we dealt with two fractional terms in parenthesis.

But if there were no parenthesis, then we would NOT use the quotient rule. Is this correct?

\(\displaystyle f(x) = \bigg[\dfrac{(x-6)}{(x+9)}\bigg]^4\)

\(\displaystyle f'(x) = 4^3\cdot\)u'\(\displaystyle \ \ \ \ \ \ \ \)Before this step, you must state what you are letting the variable u represents.

\(\displaystyle f'(x) = 4^{3}[\dfrac{(x + 9)(1) - (x - 6)(1)}{(x + 9)^{2}}]\) - \(\displaystyle u(du)\).


For \(\displaystyle du\), use the quotient rule: \(\displaystyle [g(f') - f(g')]/g^{2} \ \ \ \ \) <---- You were missing some grouping symbols in the numerator.\(\displaystyle \ \ \ \ \ \ \ \ \) Given \(\displaystyle \dfrac{f}{g}\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{(x + 9) - (x - 6)}{(x + 9)^{2}}]\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{x + 9 - x + 6}{(x + 9)^{2}}]\)

\(\displaystyle f'(x) = 4^{3}[\dfrac{3}{(x + 9)^{2}}] \ \ \ \ \ \)The numerator is incorrect. 9 + 6 = 15, not 3.

\(\displaystyle f'(x) = 4[\dfrac{(x-6)}{(x+9)}]^{3}[\dfrac{3}{(x + 9)^{2}}]\) - Back Substitute

That last step you have isn't the final form. You have to multiply all of the constants together into one constant out front.

All of the (x + 9) factors have to be consolidated into one expression.

.

If \(\displaystyle u=(x-6)/(x-9)\), then using the quotient rule \(\displaystyle u' = -3/(x-9)^2\).

/\ /\ daon2, the original denominator is x + 9.



\(\displaystyle f'(x) \ = \ 4\bigg[\dfrac{(x-6)}{(x+9)}\bigg]^{3}\bigg[\dfrac{15}{(x + 9)^{2}}\bigg]\)



\(\displaystyle f'(x) \ = \ 60\bigg[\dfrac{(x-6)^3}{(x+9)^3}\bigg]\bigg[\dfrac{1}{(x + 9)^{2}}\bigg]\)



\(\displaystyle \boxed{f'(x) \ = \ 60\bigg[\dfrac{(x-6)^3}{(x+9)^5}\bigg] \ \ \ \ or \ \ \ \ \dfrac{60(x-6)^3}{(x+9)^5} \ } \)