# Thread: Derivatives of equation using ln and e

1. ## Derivatives of equation using ln and e

Here is the equation:

Find $\displaystyle \frac{dy}{dx}$ for $\displaystyle e^ylnx+xy^2=y$

I know that

$\displaystyle f'$ of $\displaystyle e^y=e^y$

$\displaystyle f'$ of $\displaystyle lnx=\frac{1}{x}$

and

$\displaystyle f'$ of $\displaystyle xy^2=2xy$

I suppose I use the chain rule to work out the answer?

This is part of an assignment that is reviewing a few things we've covered and earlier today I listed some other problems, I have a multiple choice list of solutions that I'm posting for reference

a.) $\displaystyle \frac{-(e^y+xy^2)}{x(e^ylnx+2y-1)}$

b.) $\displaystyle \frac{-(e^y+xy)}{x(e^ylnx+2xy-1)}$

c.) $\displaystyle \frac{e^y+xy}{e^ylnx+2xy}$

d.) None of these

Thanks in advance for your help. I'm just a bit intimidated by the possibility of making a mistake in the steps

2. ## Here you go

Originally Posted by XIII13Thirteen
Here is the equation:

Find $\displaystyle \frac{dy}{dx}$ for $\displaystyle e^ylnx+xy^2=y$

I know that

$\displaystyle f'$ of $\displaystyle e^y=e^y$

$\displaystyle f'$ of $\displaystyle lnx=\frac{1}{x}$

and

$\displaystyle f'$ of $\displaystyle xy^2=2xy$

I suppose I use the chain rule to work out the answer?

This is part of an assignment that is reviewing a few things we've covered and earlier today I listed some other problems, I have a multiple choice list of solutions that I'm posting for reference

a.) $\displaystyle \frac{-(e^y+xy^2)}{x(e^ylnx+2y-1)}$

b.) $\displaystyle \frac{-(e^y+xy)}{x(e^ylnx+2xy-1)}$

c.) $\displaystyle \frac{e^y+xy}{e^ylnx+2xy}$

d.) None of these

Thanks in advance for your help. I'm just a bit intimidated by the possibility of making a mistake in the steps
$\displaystyle e^{y}\cdot\ln(x)+xy^2=y$

then $\displaystyle \frac{e^{y}}{x}+\ln(x)e^{y}\cdot{y'}+2xyy'+y^2=y'$...now solve for y'

3. Originally Posted by Mathstud28
$\displaystyle e^{y}\cdot\ln(x)+xy^2=y$

then $\displaystyle \frac{e^{y}}{x}+\ln(x)e^{y}\cdot{y'}+2xyy'+y^2=y'$...now solve for y'
I'm a bit lost. Inserting the y' into the equation looks familiar, but I think I've forgotten how this principle is applied

4. just note that $\displaystyle y' = \frac{dy}{dx}$

it came out because of your "chain rule"..

if your $\displaystyle y=e^y \ln{x} + xy^2$ and you are looking for $\displaystyle y' = \frac{dy}{dx}$.. here it is..

taking the derivative of both sides yields to
$\displaystyle \frac{dy}{dx} = \underbrace{\frac{e^y}{x} + e^y\, \ln{x} \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, e^y} + \underbrace{y^2 + 2xy \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, y^2}$

5. Originally Posted by kalagota
just note that $\displaystyle y' = \frac{dy}{dx}$

it came out because of your "chain rule"..

if your $\displaystyle y=e^y \ln{x} + xy^2$ and you are looking for $\displaystyle y' = \frac{dy}{dx}$.. here it is..

taking the derivative of both sides yields to
$\displaystyle \frac{dy}{dx} = \underbrace{\frac{e^y}{x} + e^y\, \ln{x} \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, e^y} + \underbrace{y^2 + 2xy \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, y^2}$
Ahh, thank you for clarifying. It seems that the answer is "none of these" in this case. I suppose I assumed my answer would be very close to the answers provided, so I was wondering about the denominator in particular and the negative symbols. I guess the red herring was that there was no red herring