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:


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

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




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.


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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