1. ## Implicit Differentiation question

Hi
Need help to use the Implicit Differentiation method on $\frac{xy}{lnx}$.
I have tried many ways to get the right answer. Below is what i have done:

$=\frac{\frac{dy}{dx}lnx-\frac{xy}{x}}{(lnx)^2}$

$\frac{\frac{dy}{dx}lnx}{(lnx)^2}-\frac{\frac{xy}{x}}{(lnx)^2}$

$\frac{dy}{dx}\frac{1}{lnx}=\frac{\frac{xy}{x}}{(ln x)^2}$

$\frac{dy}{dx}=\frac{y}{(ln)^2}-\frac{1}{lnx}$

$\frac{dy}{dx}=\frac{y-lnx}{(lnx)^2}$

Book's answer is $\frac{y-ylnx}{xlnx}$
P.S

2. Originally Posted by Paymemoney
Hi
Need help to use the Implicit Differentiation method on $\frac{xy}{lnx}$.
I have tried many ways to get the right answer. Below is what i have done:

$=\frac{\frac{dy}{dx}lnx-\frac{xy}{x}}{(lnx)^2}$

$\frac{\frac{dy}{dx}lnx}{(lnx)^2}-\frac{\frac{xy}{x}}{(lnx)^2}$

$\frac{dy}{dx}\frac{1}{lnx}=\frac{\frac{xy}{x}}{(ln x)^2}$

$\frac{dy}{dx}=\frac{y}{(ln)^2}-\frac{1}{lnx}$

$\frac{dy}{dx}=\frac{y-lnx}{(lnx)^2}$

Book's answer is $\frac{y-ylnx}{xlnx}$
P.S
You should apply the product rule when you differentiate $xy$.

3. ok i'll try that

4. ok after doing it again this is what i got:
$\frac{xy}{lnx}=\frac{y+x}{lnx}$

$\frac{dy}{dx}lnx-\frac{y+x}{x}$

$\frac{\frac{dy}{dx}xlnx}{x}-\frac{y+x}{x}$

$\frac{\frac{dy}{dx}xlnx-y+x}{x(lnx)^2}$

$\frac{\frac{dy}{dx}lnx}{x(lnx)^2}=\frac{y+x}{x(lnx )^2}$

$\frac{y+x-x{lnx}}{x(lnx)^2}$

5. Originally Posted by Paymemoney
ok after doing it again this is what i got:
$\frac{xy}{lnx}=\frac{y+x}{lnx}$
What is this ?

6. Originally Posted by Paymemoney
ok after doing it again this is what i got:
$\frac{xy}{lnx}=\frac{y+x}{lnx}$

$\frac{dy}{dx}lnx-\frac{y+x}{x}$

$\frac{\frac{dy}{dx}xlnx}{x}-\frac{y+x}{x}$

$\frac{\frac{dy}{dx}xlnx-y+x}{x(lnx)^2}$

$\frac{\frac{dy}{dx}lnx}{x(lnx)^2}=\frac{y+x}{x(lnx )^2}$

$\frac{y+x-x{lnx}}{x(lnx)^2}$
$\left(\frac{xy}{\ln x}\right)' =\frac{(y + xy')\ln x - xy(\frac{1}{x})} {\ln^2 x}= 0$

$(y + xy')\ln x - y= 0$

$(y + xy') = \frac{y}{\ln x}$

$xy' = \frac{y}{\ln x} - y$

$y' = \frac{y}{x\ln x} - \frac{y}{x}$

$y' = \frac{y - y\ln x}{x\ln x}$

7. Originally Posted by Paymemoney
Hi
Need help to use the Implicit Differentiation method on $\frac{xy}{lnx}$.
I have tried many ways to get the right answer. Below is what i have done:

$=\frac{\frac{dy}{dx}lnx-\frac{xy}{x}}{(lnx)^2}$

$\frac{\frac{dy}{dx}lnx}{(lnx)^2}-\frac{\frac{xy}{x}}{(lnx)^2}$

$\frac{dy}{dx}\frac{1}{lnx}=\frac{\frac{xy}{x}}{(ln x)^2}$

$\frac{dy}{dx}=\frac{y}{(ln)^2}-\frac{1}{lnx}$

$\frac{dy}{dx}=\frac{y-lnx}{(lnx)^2}$

Book's answer is $\frac{y-ylnx}{xlnx}$
P.S
What is $\frac{xy}{lnx}$ equal to???

8. Originally Posted by General
What is this ?
i used the product rule on xy to get y+x

9. omg i forgot to add that $\frac{xy}{lnx}=2$

10. Originally Posted by Paymemoney
omg i forgot to add that $\frac{xy}{lnx}=2$
1- You did not apply the product rule on $xy$ correctly.
2- Its wrong to apply the product rule on the numerator first, then differentiate the whole ratio.

11. In order to use implicit differentiation to find dy/dx, you have to have an equation to start with.

So I ask what xalk asked before: "What is $
\frac{xy}{lnx}
$
equal to?"

12. Originally Posted by Paymemoney
Hi
Need help to use the Implicit Differentiation method on $\frac{xy}{lnx}$.
I have tried many ways to get the right answer. Below is what i have done:

$=\frac{\frac{dy}{dx}lnx-\frac{xy}{x}}{(lnx)^2}$

$\frac{\frac{dy}{dx}lnx}{(lnx)^2}-\frac{\frac{xy}{x}}{(lnx)^2}$

$\frac{dy}{dx}\frac{1}{lnx}=\frac{\frac{xy}{x}}{(ln x)^2}$

$\frac{dy}{dx}=\frac{y}{(ln)^2}-\frac{1}{lnx}$

$\frac{dy}{dx}=\frac{y-lnx}{(lnx)^2}$

Book's answer is $\frac{y-ylnx}{xlnx}$
P.S
Another way of doing it is the following:

$\frac{xy}{lnx}=2\Longrightarrow xy = 2lnx$ $\Longrightarrow \frac{d(xy)}{dx} =2\frac{d(lnx)}{dx}$ $\Longrightarrow y+xy' = \frac{2}{x}$ $\Longrightarrow y' = \frac{2-xy}{x^2}$.

And if we substitute 2 with xy/lnx we have:

$y' = \frac{y-ylnx}{lnx}$