Question

I keep getting this wrong find the derivative using implicit differentiation
xe^y-ye^x=4

Answer 1

What have you tried so far?

Answer 2

\( \displaystyle x e^{y} - y e^{x} = 4 \)

We take the derivative of everything on each side.
\( \displaystyle \frac{\text{d}}{\text{d}x} \left( x e^{y} - y e^{x} \right) = \frac{\text{d}}{\text{d}x} \left( 4 \right) \)
Then, we must use some of the properties of derivatives to start tearing up this problem.

Answer 3

\( \displaystyle \frac{\text{d}}{\text{d}x} \left( x e^{y} - y e^{x} \right) = \frac{\text{d}}{\text{d}x} \left( 4 \right) \)
We have the derivative of a sum, which is equivalent to the sum of the derivatives of each term.
We also have the derivative of a constant, which will always be 0.
\( \displaystyle \color{green}{\frac{\text{d}}{\text{d}x}} \left( x e^{y} \right) - \color{green}{\frac{\text{d}}{\text{d}x}} \left( y e^{x} \right) = \color{green}{0} \)

Answer 4

\( \displaystyle \frac{\text{d}}{\text{d}x} \left( x e^{y} \right) - \frac{\text{d}}{\text{d}x} \left( y e^{x} \right) = 0 \)
Here, we now have the derivative of a product. We recall our identity here:
\( \displaystyle \color{goldenrod}{\star \star} \quad \frac{\text{d}}{\text{d}x} \left( f(x) g(x) \right) = \frac{\text{d}}{\text{d}x} \left( f(x) \right) g(x) + f(x) \frac{\text{d}}{\text{d}x} \left( g(x) \right) \)

So that we may break apart those products into individual derivatives of products. We want to be careful while breaking apart that second term since it should all be subtracted.
\( \displaystyle \frac{\text{d}}{\text{d}x} \left(x\right) e^{y} + x \frac{\text{d}}{\text{d}x} \left(e^{y}\right) - \left( \frac{\text{d}}{\text{d}x} \left( y \right) e^{x} + y \frac{\text{d}}{\text{d}x} \left( e^{x} \right) \right) = 0 \)

Answer 5

\( \displaystyle \frac{\text{d}}{\text{d}x} \left(x\right) e^{y} + x \frac{\text{d}}{\text{d}x} \left(e^{y}\right) - \left( \frac{\text{d}}{\text{d}x} \left( y \right) e^{x} + y \frac{\text{d}}{\text{d}x} \left( e^{x} \right) \right) = 0 \)

Now, we have some more identities we could use. We should know the formula for derivative of x to any power, and also the exponential function:
\( \displaystyle \color{green}{\star} \quad \frac{\text{d}}{\text{d}x} \left( x^n \right) = n x^{n-1} \)
\( \displaystyle \color{green}{\star} \quad \frac{\text{d}}{\text{d}x} \left( e^{u(x)} \right) = \frac{\text{d}}{\text{d}x} \left(u(x) \right) e^{u} = u' e^{u} \)

Now, it was a little unnecessary to restate the power rule for our simple factor of x, but it's always good to remember those identities. x = x^1, so 1*x^(1-1) = 1 * x^0 = 1. Let's continue simplifying now:
\( \displaystyle \color{red}{1} e^{y} + x \color{red}{\frac{\text{d}}{\text{d}x} \left(y \right) e^{y} } - \left( \color{red}{1*\frac{\text{d}y}{\text{d}x}} e^{x} + y \color{red}{e^{x}} \right) = 0 \)

Answer 6

\( \displaystyle e^{y} + x \frac{\text{d}y}{\text{d}x} e^{y} - \frac{\text{d}y}{\text{d}x} e^{x} - y e^{x} = 0 \)
Cleaning up some things and distributing the negative.

Answer 7

Now, we want the derivative of y with respect to x isolated. So we can move anything without that factor to the right-hand side, and the remaining terms will stay on the left.
\( \displaystyle x \frac{\text{d}y}{\text{d}x} e^{y} - \frac{\text{d}y}{\text{d}x} e^{x} = y e^{x} - e^{y} \)
This allows us to factor out the derivative entirely.
\( \displaystyle \frac{\text{d}y}{\text{d}x} \left( xe^{y} - e^{x} \right) = y e^{x} - e^{y} \)
Now, we simply must divide off that other factor to isolate the derivative.
\( \displaystyle \frac{\text{d}y}{\text{d}x} = \frac{y e^{x} - e^{y}}{xe^{y} - e^{x}} \)