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I think I am stuck here right now:

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

enter image description here

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...

enter image description here

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:

enter image description here

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  • $\begingroup$ You'll need to replace e (lower case) with E (upper case). But I'm not sure what you are actually trying to do. $\endgroup$ – bill s Jun 13 at 20:24
  • $\begingroup$ 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. $\endgroup$ – Roman Jun 13 at 21:06
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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 $y'(x)=y/x$.

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

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