# Thread: Chain Rule Leibniz Notation

1. ## Chain Rule Leibniz Notation

Hi, I have been stuck on this question for a really long time any help would be appreciated.

Using leibniz notation, apply the chain to determine (dy/dx) at the indicated value of x.

$\displaystyle y= u(2-u^2), u= (1/x), x=2$

I think that I have to use the product rule and then thee chain rule but I'm lost.

2. Originally Posted by darksoulzero
Hi, I have been stuck on this question for a really long time any help would be appreciated.

Using leibniz notation, apply the chain to determine (dy/dx) at the indicated value of x.

$\displaystyle y= u(2-(u^2)), u= (1/x), x=2$

I think that I have to use the product rule and then thee chain rule but I'm lost.
First of all this is very hard to read.

I assume that it's

$\displaystyle y = u(2 - u^2)$ and $\displaystyle u = \frac{1}{x}$, where $\displaystyle x = 2$.

You need to find $\displaystyle \frac{du}{dx}$ and $\displaystyle \frac{dy}{du}$.

You are correct that to find $\displaystyle \frac{dy}{du}$ you should use the product rule. Another alternative is to expand the expression first.

You should also note that $\displaystyle \frac{1}{x} = x^{-1}$.

Once you have those derivatives, multiply them together to get $\displaystyle \frac{dy}{dx}$ and then evaluate it at $\displaystyle x = 2$.

3. Originally Posted by Prove It
First of all this is very hard to read.

I assume that it's

$\displaystyle y = u(2 - u^2)$ and $\displaystyle u = \frac{1}{x}$, where $\displaystyle x = 2$.

You need to find $\displaystyle \frac{du}{dx}$ and $\displaystyle \frac{dy}{du}$.

You are correct that to find $\displaystyle \frac{dy}{du}$ you should use the product rule. Another alternative is to expand the expression first.

You should also note that $\displaystyle \frac{1}{x} = x^{-1}$.

Once you have those derivatives, multiply them together to get $\displaystyle \frac{dy}{dx}$ and then evaluate it at $\displaystyle x = 2$.
ok I expanded it and got $\displaystyle \frac{dy}{dx} = 2u-u^3$ now how would I get the derivative in leibniz notation.

4. Originally Posted by darksoulzero
ok I expanded it and got $\displaystyle \frac{dy}{dx} = 2u-u^3$ now how would I get the derivative in leibniz notation.
No that doesn't look right.

You have $\displaystyle y = 2u - u^3$.

So now apply the power rule to each term to find $\displaystyle \frac{dy}{du}$.

5. Originally Posted by Prove It
No that doesn't look right.

You have $\displaystyle y = 2u - u^3$.

So now apply the power rule to each term to find $\displaystyle \frac{dy}{du}$.
ok I think I got it.

$\displaystyle =2u-u^3$
$\displaystyle =2-3u^3$
$\displaystyle =2-3(-x^{-2})^2$
$\displaystyle =2-\frac{3}{-x^4}$
$\displaystyle =2-\frac{3}{-(2)^4}$

6. No.

$\displaystyle y = 2u - u^3$

$\displaystyle \frac{dy}{du} = 2 - 3u^2$

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

$\displaystyle = 2 - \frac{3}{x^2}$.

$\displaystyle u = \frac{1}{x}$

$\displaystyle = x^{-1}$.

$\displaystyle \frac{du}{dx} = -x^{-2}$

$\displaystyle = -\frac{1}{x^2}$.

$\displaystyle \frac{dy}{dx} = \frac{du}{dx}\,\frac{dy}{du}$

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

$\displaystyle = \frac{3}{x^4} - \frac{2}{x^2}$.