Implicit Differentiation of the following equation

Question

$$4x^3 + x^2y - xy^3 = 4$$

This is what I have so far:

$$(2xy + x^2 y') - (y^3 + 3xy^2 y') = -12x^2$$

Should I bring everything but the y primes over to the right side by dividing it? I'm not so sure on what to do in this situation.

Answer 1

One cool thing about implicit differentiation of $y$ via an expression such as

$f(x, y) = 0 \tag{1}$

is that the resulting derivative, i.e.,

$\frac{\partial f}{\partial x} + y'(x)\frac{\partial f}{\partial y} = 0, \tag{2}$

is always a linear expression in the function $y'(x)$, and as such, is generally easy to solve for $y'(x)$; indeed, from (2),

$y'(x) = -\frac{\partial f}{\partial x}(\frac{\partial f}{\partial y})^{-1}, \tag{3}$

as long as $\frac{\partial f}{\partial y} \ne 0$. Applying these notions to the case at hand,

$(2xy + x^2 y') - (y^3 + 3xy^2 y') = -12x^2, \tag{4}$

we see it can be re-arranged to give

$(2xy - y^3) + y'(x^2 - 3xy^2) = -12x^2, \tag{5}$

an expression linear in $y'$, for which it is easily solved:

$y'(x) = \frac{y^3 -2xy -12x^2}{x^2 - 3xy^2}, \tag{6}$

which allows $y'(x)$ to be calculated once $x$ and $y$ satisfying $f(x, y) = 0$ are known (here $f(x, y) = 4x^3 + x^2y - xy^3 - 4$).

Hopes this helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 2

Rather than using $y^\prime$ etc., you might simply want to use the differential operator as $df$, $dx$, $dy$, etc.:

$$d(4x^3 + x^2y - xy^3) = d(4)=0$$

That is,

$$12x^2dx + 2xydx + x^2dy - \cdots =0$$

For example, here $d(x^2y)$ becomes $2xydx + x^2dy $ by the product and chain rules.

Now, group the above as

$$(\operatorname{Expression 1})dx = (\operatorname{Expression 2})dy $$

And from this you get $dy/dx$.

This is a more convenient way to write down calculations of implicit differentiation, once you get the hang of it.