Solve for $\frac{dy}{dx}$ of $\ln(2xy)=e^{x+y}$

[karush]

Find $\frac{dy}{dx}$

$$\ln\left({2xy}\right)=e^{x+y}$$

ok i tried to get both sides either $\ln$ or $e$ but couldn't

 

[MarkFL]

Find $\frac{dy}{dx}$

$$\ln\left({2xy}\right)=e^{x+y}$$

ok i tried to get both sides either $\ln$ or $e$ but couldn't

What you want to do here is implicitly differentiate both sides w.r.t $x$:

\(\displaystyle \frac{1}{2xy}\left(2\left(y+x\d{y}{x}\right)\right)=e^{x+y}\left(1+\d{y}{x}\right)\)

Now solve for \(\displaystyle \d{y}{x}\)...:)

 

[karush]

really? I'm lost!

Is the answer btw? (this was from multiple choice)

$$\displaystyle\frac{xye^{x+y}-y}{x-xye^{x+y}}$$

 

[MarkFL]

really?

I'm lost!

Where did I lose you? (Wondering)

 

[karush]

the left side how do you $dy/dx$ a nested factor

 

[MarkFL]

the left side how do dy/dx a nested factor

I'm not sure I know what you mean by that, but on the LHS, I applied the rule for differentiating the natural log function, along with the chain and product rules. :)

 

[MarkFL]

Is the answer btw? (this was from multiple choice)

$$\displaystyle\frac{xye^{x+y}-y}{x-xye^{x+y}}$$

Let's go back to:

\(\displaystyle \frac{1}{2xy}\left(2\left(y+x\d{y}{x}\right)\right)=e^{x+y}\left(1+\d{y}{x}\right)\)

Divide out the common factor on the LHS:

\(\displaystyle \frac{1}{xy}\left(y+x\d{y}{x}\right)=e^{x+y}\left(1+\d{y}{x}\right)\)

Multiply through by $xy$:

\(\displaystyle y+x\d{y}{x}=xye^{x+y}\left(1+\d{y}{x}\right)\)

Distribute on the RHS:

\(\displaystyle y+x\d{y}{x}=xye^{x+y}+xye^{x+y}\d{y}{x}\)

Arrange with all terms having \(\displaystyle \d{y}{x}\) as a factor on the LHS, and everything else on the RHS:

\(\displaystyle x\d{y}{x}-xye^{x+y}\d{y}{x}=xye^{x+y}-y\)

Factor:

\(\displaystyle x\left(1-ye^{x+y}\right)\d{y}{x}=y\left(xe^{x+y}-1\right)\)

Divide through by \(\displaystyle x\left(1-ye^{x+y}\right)\):

\(\displaystyle \d{y}{x}=\frac{y\left(xe^{x+y}-1\right)}{x\left(1-ye^{x+y}\right)}\quad\checkmark\)

 

[karush]

wow
much mahalo

that was a tough one