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Please excuse if the formatting of this post is wrong.

There's a question that asks for the 2nd derivative of $y-2x-3xy=2$

From what I know, I have to use implicit differentiation, using which I get: $$\frac{12+18y}{(1-3x)^{2}}$$ But can you solve for y in the initial equation and differentiate two times (aka explicit differentiation)? By doing that I got: $$\frac{48}{(1-3x)^{3}}$$ I'm not sure if this is a correct answer as I am new to differentiation. I guess that this question is also tied with another question; can you substitute y in implicit differentiation by solving for it in the initial equation?

Thank you.

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    $\begingroup$ The derivative is defined for a function, not an equation. I don't know what "the derivative of y- 2x+ 3xy= 2" mean. If you mean the derivative of y with respect to x, then y+ 3xy= y(1+ 3x)= 2+ 2x so $y= \frac{2(1+ x)}{1+ 3x}= 2(1+ x)(1+ 3x)^{-1}$. $\endgroup$
    – user247327
    Commented Nov 24, 2020 at 14:02

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Sure you can in this case $$y=\frac{2 (x+1)}{1-3 x}$$ and then $$y'=\frac{8}{(1-3 x)^2}$$ and finally $$y''=\frac{48}{(1-3 x)^3}$$

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  • $\begingroup$ Is there any situation where using this is not correct? Maybe like x^2+y^2=25? $\endgroup$
    – MHDev
    Commented Nov 24, 2020 at 12:57
  • $\begingroup$ @MHDev In your example you can't solve for $y$ and get a function. You get two values. Even worse if you had to differentiate $x^3+y^3=\sin(xy)$. You can do like I did only if you can solve wrt $y$ or wrt $x$ $\endgroup$
    – Raffaele
    Commented Nov 24, 2020 at 13:05
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(This was too long to be a comment.)

You can substitute $ \ y \ $ into the implicit differentiation calculation when it is a single function suggested by the original equation. Often, though, that equation may be describing two or more implicit functions and it is not necessarily straightforward to select the correct one for substitution.

You can get away with it in this problem since there is just one explicit function, $ \ y \ = \ \frac{2·(1 \ + \ x)}{1 \ - \ 3x} \ \ . $ Inserting that into your first result gives $$ y'' \ \ = \ \ \frac{12 \ + \ 18y}{(1 \ - \ 3x)^2 } \ \ = \ \ \frac{12 \ + \ 18·\left[ \ \frac{2·(1 \ + \ x) }{1 \ - \ 3x} \ \right]}{(1 \ - \ 3x)^2} \ · \ \frac{1 \ - \ 3x}{1 \ - \ 3x} $$ $$ = \ \ \frac{12 · (1 \ - \ 3x) \ + \ 18· 2·(1 \ + \ x) }{(1 \ - \ 3x)^3 } \ \ = \ \ \frac{12 \ - 36x \ + \ 36 \ + \ 36x }{(1 \ - \ 3x)^3 } \ \ = \ \ \frac{12 · (1 \ - \ 3x) \ + \ 18· 2·(1 \ + \ x) }{(1 \ - \ 3x)^3 } \ \ = \ \ \frac{48}{(1 \ - \ 3x)^3 } \ \ . $$

So you were doing fine here. Since this is an old question, you likely soon found for other implicit differentiation problems that the avenues for substitution were far more limited.

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We differentiate once,

$$y'-2-3y-3xy'=0$$ and twice,

$$y''-3y'-3y'-3xy''=0.$$

Now we can eliminate $y'$, using

$$(1-3x)y'=3y+2$$

and

$$(1-3x)y''=6y'=6\frac{3y+2}{1-3x}.$$

You can also eliminate $y$ using the initial equation.

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  • $\begingroup$ I have corrected a copying error from the original equation: the end result now agrees with OP's calculation of $ \ y'' \ \ . $ $\endgroup$
    – user882145
    Commented Mar 11, 2022 at 9:43

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