Can you simplify an equation to use explicit differentiation instead of implicit differentiation?

Question

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.

Answer 1

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 $$y''=\frac{48}{(1-3 x)^3}$$

Answer 2

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.

Answer 3

(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.