# Implicit Differentiation question

• Feb 11th 2010, 09:47 PM
Paymemoney
Implicit Differentiation question
Hi
Need help to use the Implicit Differentiation method on $\displaystyle \frac{xy}{lnx}$.
I have tried many ways to get the right answer. Below is what i have done:

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

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

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

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

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

Book's answer is $\displaystyle \frac{y-ylnx}{xlnx}$
P.S
• Feb 11th 2010, 09:54 PM
General
Quote:

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

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

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

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

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

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

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

You should apply the product rule when you differentiate $\displaystyle xy$.
• Feb 11th 2010, 09:57 PM
Paymemoney
ok i'll try that
• Feb 12th 2010, 01:04 AM
Paymemoney
ok after doing it again this is what i got:
$\displaystyle \frac{xy}{lnx}=\frac{y+x}{lnx}$

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

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

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

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

$\displaystyle \frac{y+x-x{lnx}}{x(lnx)^2}$
• Feb 12th 2010, 01:18 AM
General
Quote:

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

What is this ?
• Feb 12th 2010, 01:20 AM
dedust
Quote:

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle y' = \frac{y - y\ln x}{x\ln x}$
• Feb 12th 2010, 01:22 AM
xalk
Quote:

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

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

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

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

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

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

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

What is $\displaystyle \frac{xy}{lnx}$ equal to???
• Feb 12th 2010, 01:27 AM
Paymemoney
Quote:

Originally Posted by General
What is this ?

i used the product rule on xy to get y+x
• Feb 12th 2010, 01:34 AM
Paymemoney
omg i forgot to add that $\displaystyle \frac{xy}{lnx}=2$
• Feb 12th 2010, 01:38 AM
General
Quote:

Originally Posted by Paymemoney
omg i forgot to add that $\displaystyle \frac{xy}{lnx}=2$

1- You did not apply the product rule on $\displaystyle xy$ correctly.
2- Its wrong to apply the product rule on the numerator first, then differentiate the whole ratio.
• Feb 12th 2010, 02:04 AM
HallsofIvy
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 $\displaystyle \frac{xy}{lnx}$ equal to?"
• Feb 12th 2010, 02:18 AM
xalk
Quote:

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

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

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

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

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

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

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

Another way of doing it is the following:

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

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

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