# Implicit Function Theorem

I think I am stuck here right now:

The exercise is to use implicit differentiation to determine y' considering the following equation:

I managed to do this with:

eqn = e^(a*x/y[x]) + e^(b*x/y[x]) == c;
Solve[D[eqn, x], y'[x]]


which lead to the result below, which should be correct...

The next exercise is to verify the solution with the implicit function theorem, but I have no clue how to do that. Would appreciate every answer :D

Edit: Changing e to E does not make any difference

Here is the whole instruction:

• You'll need to replace e (lower case) with E (upper case). But I'm not sure what you are actually trying to do. – bill s Jun 13 at 20:24
• Just staring at your equation, I'd guess that $x/y$ must be a constant depending on $\{a,b,c\}$: let's call $x/y=f$, or $y=x/f$. From this you get $dy/dx=1/f$, which is constant as well. Figuring out $f$ is tricky, as Solve[E^(a*f) + E^(b*f) == c, f] does not return a solution. – Roman Jun 13 at 21:06

First implicitly differentiate the expression with respect to x: $$\frac{d}{dx}\{e^{ax/y}+e^{bx/y}=c\}$$ We can do this in Mathematica:

 myImplicitDeriv=D[Exp[a x/y[x]] + Exp[b x/y[x]] ==
c, x]


Now solve for y'[x]:

myDeriv=Solve[myImplicitDeriv /. {y'[x] -> y', y[x] -
>y}, y']


with the results being $$y'[x]=y/x$$. This is the implicit part. Now, the Implicit Function Theorem states for $$f(x,y)=0$$, we can determine y'(x) as:

$$y'(x)=-\frac{f_x}{f_y}$$

So then let:

$$f(x,y)=e^{ax/y}+e^{bx/y}-c=0$$ then write:

f[x_, y_] = Exp[a x/y] + Exp[b x/y] - c
Simplify[-D[f[x, y], x]/D[f[x, y], y]]


which also yields $$y'(x)=y/x$$.

• Thank you very much, that's exactly what I was looking for :D – stefan Jun 14 at 15:50