Question

I need assistance in finding...
dy/dx of the equation: "y = (x/x-1)^(-3)"
ANY help is greatly appreciated! :)

Answer 1

@agent0smith Buddy ole' pal... (?) :)

Answer 2

i'd start with
\[y=\left(\frac{x-1}{x}\right)^3\]

Answer 3

Haha. Alright... I already need explanation. What's you do there?

Answer 4

\[\Large y=\left(\frac{x-1}{x}\right)^3 = \]algebra will be your friend\[\Large y =(1 - x^{-1})^3\]Use chain rule

Answer 5

Alright... I believe I know where to go. I'm gonna work through it and post what I get if that's alright. (?)

Answer 6

So I got:
y = [ 3((x-1)/x)^2] + ... But I don't know how to take the next derivative of the inside.

Answer 7

Like I said, algebra is your friend \[\Large y =(1 - x^{-1})^3\]

Answer 8

I don't even understand satelite73's post, at the top of this question... What did he do, in concrete terms?

Answer 9

How did the equation you posted come from my original one?

Answer 10

\[\Large \left( \frac{ a }{ b} \right)^{-c} = \left( \frac{ b }{ a} \right)^{c}\]

Answer 11

Exponent rules^^

Then what I did was just simplify the fraction.

Answer 12

Okay, so even if you're not just putting one term into the denominator it's still the concept ?

Bottom to top= exponent sign switch
+
Top to bottom = exponent sign switch

Knowing this brings immense clarity. Alright... So,

(x/x-1)^(-3) = (x-1/x)^3 = [(x/x) - (1/x)]^3 = [ 1 - (x^-1)]^3
Did I get through all of that correctly?

Answer 13

Yes.

Answer 14

Do I then make it as follows before taking the derivative:

y = x^3 - (x^-3)
...by distributing the cubed (3) inside, and multiplying it by the power of (-1) in the last term ?

Answer 15

No, you can distribute the cubed inside like that. \[\Large \frac{ d }{ dx}\left( f(x) \right)^a = a \left( f(x) \right)^{a -1}*f'(x)\]

Answer 16

*can't distribute the cubed like that

Answer 17

So...

y = {x^3 - [x^(-1)]^3}
y' = {(3x^2) - [3x^(-2)]} * [2x^(2-1)] * [-2x^(-2-1)]
= {3x^2 - 3x^(-2)} * 2x * (-2x)^(-3)
= {3x^2 - 3x^(-2)} * (-4x^2)^(-3)

Good so far? :/

Answer 18

The function inside the parentheses shouldn't have changed, see the formula above.

Answer 19

\[\Large \frac{ d }{ dx}\left( f(x) \right)^a = a \left( f(x) \right)^{a -1}*f'(x)\]
here, f(x) is \[\large (1-x^{-1})\]and a is 3

Answer 20

And again, y = {x^3 - [x^(-1)]^3} is not correct, can't "distribute" the exponent like that. It's \[y = \large (1-x^{-1})^3\]

Answer 21

Alright, thank you for your help! :)