Question

Use implicit differentiation to find dy/dx where x^2y=1 + e^y

Answer 1

Do you know how to do Implicit D?

Answer 2

A little bit. But I'm confused with the e^y

Answer 3

Chain rule it.

\[\left(e^{u(x)}\right )^{\prime} = e^{u(x)} (u(x))^{\prime}\] right?
just replace u(x) with y(x)

Answer 4

I have never seen that before... I'm familiar with the chain rule but have never seen it done like that.

Answer 5

So they probably taught you the chain rule by
\[\frac{d}{dx}\bigg(stuff(x)\bigg)^7= 7\bigg(stuff(x)\bigg)^6\]
or something like that?

Answer 6

yeah exactly

Answer 7

Yeah its easier to understand what to do that way but harder on explaining Implicit D.

Notice that I actually finished wrong it should have been
\[7\bigg(stuff(x)\bigg)^6 \bigg(stuff(x) \bigg)^{\prime}\]
right?
the issue here is that you are taught basic derivatives like
\[ f(x) = x^4\to f^{\prime}(x) = 4x^3\]
So what happens when you have
\[f(x) = (g(x))^4\]
It looks like

\[ f(x) = x^4\to f^{\prime}(x) = 4x^3\]
but who knows what g(x) is.
So they came up with this cool fact that
\[\frac{df}{dx}=\frac{df}{dg}\frac{dg}{dx}\]
Notice that
\[\frac{df}{dg} = 4(g(x))^3\]
and
\[\frac{dg}{dx}= (g(x))^{\prime}\]
It basically mimics the stuff argument but with a variable for "stuff(x)"
So instead of actually finding the (g(x))' you leave it for implicit D and then factor and solve for it.

Answer 8

I understand it a little bit more, but I'm still so confused.

Answer 9

Do you know of any good websites or books that can help me with that? My textbook is very confusing.

Answer 10

just math tutoring has a video on this that makes it look painfully easy
the explanation is something that will make a math teacher cringe, but it works
i will see if i can find it

Answer 11

unfortunately i cannot link to the video exactly
but it is here, just scroll down to "calculus" and then "implicit differentiation"
http://justmathtutoring.com/

Answer 12

here's a PDF I wrote. It gives an example and reviews the chain rule.

Answer 13

justmathtutoring aka patrickjmt.com? yeah i use his site a lot, but i didn't see a problem with e^y

Answer 14

okay thanks guys. i'll keep at it.

Answer 15

Remember that e^y is just a function that takes y as a variable. you could argue that its
e(y) or as it is sometimes written exp(y).

Answer 16

just like cos(y)