Question

Find dy/dx implicitly.

Answer 1

\[e^{xy}+x ^{2}-y ^{2}=10\]

Answer 2

@.Sam. @abb0t

Answer 3

What would implicit differentiation give yo for the 2nd and 3rd term on the left side?

Answer 4

2x-2y(dy/dx)=0

Answer 5

just have trouble with the e

Answer 6

OK.

To find the derivative of e to the anything you must first write down the same e to the anything.

In your case simply rewrite e^xy.

Then, by the chain rule, take the derivative of the exponent, and multiply it by e^ xy.

What do you get?

Answer 7

How would you do the chain rule in this case? I keep thinking to use product rule between the x and y...

Answer 8

You need both the chain rule AND the product rule.

d/dx e^xy = e^xy times (d/dx xy) That's the chain rule

Using the product rule, find d/dx of xy.

Answer 9

d/dx(x)+y.

but how did you get that for the chain rule? i'm extremely confused as to how the chain rule works when you have e

Answer 10

OK. The chain rule applies when one has a function OF a function.

Suppose you have the derivative of e^2x. First you rewrite e^2x, then multiply by the derivative of the exponent (in this case, 2) So d/dx e^2x = 2e^2x.

With me?

Answer 11

Yes.

Answer 12

OK. What's the derivative of \[e ^{3x ^{2}}\]

Answer 13

That has nothing to do with the specifics of your question, I just want to take it easy.

Answer 14

\[e ^{3x ^{2}}*6x\]

Answer 15

Perfect.

Answer 16

That covers the chain rule. On to the product rule!

Now we need to multiply by the derivative of \[e ^{xy}\]

It must be \[(e ^{xy})\frac{ d }{ dx }xy\]

Answer 17

oohhh ok. so using the chain rule it's now that. ok i see it. then i use product rule within that d/dx?

Answer 18

Yes.

\[e ^{xy}(y +xy') + 2x-2yy'=0\]

I think you've got it.

I've got to go - good luck

Answer 19

thanks so much for your time and patience!