Implicit differentiation...finding equation of the tangent

Question

x^2e^y+ye^x=4 (2,0) 

 

I got y=-x+2 for the equation. Is this right ? 

Answer 1

Hey Alex.

 

I have a weird way of doing implicit differentiation but it's a method that's worked on 100% of my students. First take d/dx of the entire equation.

 

d/dx(x²*e^y + y*e^x = 4)

 

You have a couple of products here, meaning you'll need to use the product rule. My method dictates that when the d/dx comes across a variable and acts on it via differentiation, insert that variable into the numerator next to the "lonely d." I've used the analogy that the d in the numerator "is lonely by himself and just wants a friend, so whenever it comes across a variable, make sure to give it to him via the chain rule."

 

When we consider the first term, the derivative will look like this using my method:

 

d/dx(x²*e^y)

 

= 2x*(dx/dx)*e^y + x²*e^y*dy/dx

 

When d/dx acted upon x², it differentiated it to 2x. And because it came across and acted upon x, I inserted x into the numerator next to the lonely d, making dx/dx.

 

When d/dx acted upon e^y, it differentiated it to e^y. Again, because it acted upon the variable y, I gave the lonely numerator y as a friend, making dy/dx.

 

The second term has the same methodology.

 

d/dx(y*e^x) = 1*dy/dx*e^x + y*e^x*dx/dx

 

y differentiated to 1, leaving behind dy/dx.

 

e^x differentiated to e^x, leaving behind dx/dx.

 

The constant 4 just goes to 0. You're now left with:

 

2x*dx/dx*e^y + x²*e^y*dy/dx + dy/dx*e^x + y*e^x*dx/dx = 0

 

If you think of dx/dx as a fraction, you can see that it reduces to 1. So every dx/dx in the equation vanishes because multiplying by 1 is the same as leaving the term as is. dy/dx can't be reduced, so it's left alone.

 

2x*e^y + x²*e^y*dy/dx + dy/dx*e^x + y*e^x = 0

 

You may be asking yourself why I threw this step in here if it's just going to disappear to 1 anyway. When you move on from basic fundamentals of implicit differentiation into the topic of "Related Rates," wherein you'll be implicitly differentiating with respect to time (d/dt) and not x (d/dx), I'm sure you'll appreciate this method. If you're missing a dy/dt or a dx/dt in your related rates problem (because you forgot to give the lonely d up top a variable friend) it could cause lots of headaches.

 

Anyways...

 

Solving for dy/dx yields:

 

dy/dx = (-2x*e^y - y*e^x) / (x²*e^y + e^x)

 

When you plug in your point at (2,0) you get

 

dy/dx(2,0) = (-2*2*1 - 0*e²) / (4*1 + e²)

 

dy/dx(2,0) = (-4) / (4+e²)

 

So your slope at (2,0) will be

 

m = -4/(4+e²)

 

Meaning the equation of your line will be

 

y = -4/(4+e²) * (x-2)

 

I'd just leave it at that. No need to simplify it further. Wondering where the e² went in your solution? Did you forget to differentiate the y so when you plugged in 0 the term vanished??

 

Hope this helps. Lemme know if you have any questions.