very difficult derivative

[logistic_guy]

Find the derivative of $\displaystyle \left(\frac{x + 2}{x - 3}\right)^3$.

 

[khansaheb]

Find the derivative of $\displaystyle \left(\frac{x + 2}{x - 3}\right)^3$.

Use the corollary:

$\displaystyle \frac{d}{dx}{\frac{u}{v}} \ = \ \dfrac{u' * v - u \ * \ v'}{v^2}$

and

$\displaystyle \frac{d}{dx}u^{n} = n * u^{(n-1)} \ * \ u'$

 

[fresh_42]

If you can't memorize which term in @khansaheb 's first formula carries the minus sign (like me), then you can also use the Leibniz rule
$$(f(x)\cdot g(x))'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$$ and the second formula which holds for all $n\in \mathbb{Z}.$

 

[jonah2.0]

Beer induced reaction follows.

Use the corollary:

$\displaystyle \frac{d}{dx}{\frac{u}{v}} \ = \ \dfrac{u' * v - u \ * \ v'}{v^2}$

and

$\displaystyle \frac{d}{dx}u^{n} = n * u^{(n-1)} \ * \ u'$

If you can't memorize which term in @khansaheb 's first formula carries the minus sign (like me), then you can also use the Leibniz rule
$$(f(x)\cdot g(x))'=f'(x)\cdot g(x)+f(x)\cdot g'(x)$$ and the second formula which holds for all $n\in \mathbb{Z}.$

I get the impression that Mario will dazzle us with his solution very shortly along with claims of seeing this very difficult derivative for the very first time.

 

[logistic_guy]

Thanks guys for the help as I was struggling for $\displaystyle 3$ months to solve this problem!



The title of this thread is a little bit vague as I really meant very difficult derivative $\displaystyle \rightarrow$ for beginners!

Let $\displaystyle y = \left(\frac{x + 2}{x - 3}\right)^3$

Then,

$\displaystyle \frac{dy}{dx} = 3\left(\frac{x + 2}{x - 3}\right)^2\frac{x - 3 - (x + 2)}{(x - 3)^2} = \frac{-15(x + 2)^2}{(x - 3)^4}$

I don't know what I would have done without you.

 

[fresh_42]

I don't know what I would have done without you.

That's easy. There is always WA to ask about such basic questions.

y(x)=((x+2)/(x-3))^3 - Wolfram|Alpha

Wolfram|Alpha brings expert-level knowledge and capabilities to the broadest possible range of people—spanning all professions and education levels.

www.wolframalpha.com

I'm not saying you should cheat by looking it up beforehand, but you can use it to check your answer. The two formulas in post #2 and/or the Leibniz rule in post #3 should be learned by you if you want to differentiate functions. They come directly after the definition of a derivative in any calculus syllabus.

 

[khansaheb]

The two formulas in post #2 and/or the Leibniz rule

If you use that - then task will become simple and cannot use -

 

[fresh_42]

If you use that - then task will become simple and cannot use -

We should have added linearity and the chain rule to complete the list.