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Given is $F(x,y)=ye^{3x}-2x^2=0$

I was asked to calculate $y’$ using implicit differentiation.

I know that $y’=-\frac{Fx}{Fy}=-\frac{\frac{∂F}{∂x}}{\frac{∂F}{∂y}}.$

So I obtained: $y’(x)=-\frac{3ye^{3x}-4x}{e^{3x}} = \frac{-3ye^{3x}+4x}{e^{3x}}.$

But, in the solution manual I found another approach:

$y=f(x)$

i.e. $ f(x)e^{3x}-2x^2=0 $, We use the product rule

$\Rightarrow f'(x)e^{3x}+f(x)3e^{3x}-4x=0$

$f'(x)e^{3x}+f(x)3e^{3x}=4x$

$f'(x)e^{3x}=4x-(f(x)3e^{3x})$

$f'(x)=\frac{4x-(3e^{3x}f(x)}{e^{3x}}$

And finally

$ f'(x)=y'(x)=\frac{4x-3e^{3x}y}{e^{3x}} = \frac{-3ye^{3x}+4x}{e^{3x}} $

So, the result is the same.

Are both approaches valid? Is there any difference between them? Which one would you recommend to use?

Thank you in advance.

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The two approaches are indeed equivalent. In the general case, you have an equation $$ F(x, f(x)) = 0$$ where $y = f(x)$.

Differentiating both sides with respect to $x$, you get

$$ \frac{\partial F}{\partial x}(x, f(x)) + f'(x) \frac{\partial F}{\partial y}(x, f(x)) = 0, $$

and solving for $f'(x)$, you have

$$ f'(x) = - {\frac{\frac{\partial F(x,f(x))}{\partial x} }{\frac{\partial F(x,f(x))}{\partial y}}},$$ which is precisely the implicit function theorem formula that you quoted originally.

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