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

  • MHB
  • Thread starter karush
  • Start date
In summary: !in summary, the left side is differentiating $\ln\left({2xy}\right)\) and on the right side you are doing the same thing, but dividing through by x\left(1-ye^{x+y}\right) to get the dy/dx.
  • #1
karush
Gold Member
MHB
3,269
5
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
 
Physics news on Phys.org
  • #2
karush said:
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 \(\displaystyle \d{y}{x}\)...:)
 
  • #3
really? I'm lost!

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

$$\displaystyle\frac{xye^{x+y}-y}{x-xye^{x+y}}$$
 
Last edited:
  • #4
karush said:
really?

I'm lost!

Where did I lose you? (Wondering)
 
  • #5
the left side how do you $dy/dx$ a nested factor
 
Last edited:
  • #6
karush said:
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. :)
 
  • #7
karush said:
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 \(\displaystyle x\left(1-ye^{x+y}\right)\):

\(\displaystyle \d{y}{x}=\frac{y\left(xe^{x+y}-1\right)}{x\left(1-ye^{x+y}\right)}\quad\checkmark\)
 
  • #8
wow
much mahalo

that was a tough one
 

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

1. What is the general method for solving for $\frac{dy}{dx}$ in a logarithmic equation?

The general method for solving for $\frac{dy}{dx}$ in a logarithmic equation is to use the chain rule. This means that we take the derivative of the entire equation with respect to $x$, and then multiply by the derivative of the inside function, in this case $2xy$.

2. How do I isolate $\frac{dy}{dx}$ in this equation?

To isolate $\frac{dy}{dx}$ in this equation, we begin by taking the derivative of both sides using the chain rule. Then, we can manipulate the equation algebraically to solve for $\frac{dy}{dx}$. In this case, we would divide both sides by $2xy$ and then multiply by $e^{x+y}$.

3. Is it possible to solve for $\frac{dy}{dx}$ without using the chain rule?

No, it is not possible to solve for $\frac{dy}{dx}$ in a logarithmic equation without using the chain rule. This is because the chain rule allows us to take the derivative of the inside function, which is necessary in order to solve for $\frac{dy}{dx}$.

4. Can I solve for $\frac{dy}{dx}$ using any other methods?

Yes, there are other methods that can be used to solve for $\frac{dy}{dx}$ in a logarithmic equation. One alternative method is to use implicit differentiation, where we treat $y$ as a function of $x$ and take the derivative of both sides with respect to $x$. However, the chain rule is still necessary in this method.

5. What should I do if I encounter a logarithmic equation with multiple variables?

If you encounter a logarithmic equation with multiple variables, such as $\ln(2xy)=e^{x+y}$, the process for solving for $\frac{dy}{dx}$ is the same. You would still use the chain rule and algebraic manipulation to isolate $\frac{dy}{dx}$. The only difference is that you may have to use the product rule or quotient rule in addition to the chain rule.

Similar threads

Replies
3
Views
1K
Replies
4
Views
3K
Replies
4
Views
2K
Replies
5
Views
1K
Replies
3
Views
2K
Back
Top