Differentiate an equation

Question

I'm learning about implicit differentiation and i'm stuck at understanding the concept of "differentiating an equation".

If I define $ f(x) = x^2$ and $g_1(y) = 9$ and $g_2(y) = 9 + y^2$

Then I define 2 equations:
$(1)$ $f(x) = g_1(y)$ (i.e. $x^2 = 9$)
Then I can't differentiate both side of this equation with respect to x (ortherwise I will get: $2x = 0$)

$(2)$ $f(x) = g_2(y)$ (i.e. $x^2 = 9 + y^2$)
Then I can differentiate both side of this equation with respect to $x$ to get: $ 2.x = 2.y.\frac{dy}{dx}$

Could you please explain me the difference between the 2 cases ? And in general, in which condition could I differentiate both side of an equation concerning more than 1 variable ?

Thank you very much for your help!

Answer 1

$x^2=9$ is an equation that has a finite number of real solutions ($2$) for $x$. Differentiating this would not make much sense.

An equation like $x^2-1 = (x-1)(x+1)$ holds true for all real $x$. You can differentiate this and both sides will turn out to be equal.

In general, you can differentiate both sides of the equation $f_1(x,y)=f_2(x,y)$ iff this equation holds true for all real $x,y$ in a certain interval of the real numbers, containing infinitely many points.

Answer 2

When we write $x^2=9$, $x$ is an unknown, rather than a variable. By contrast, the expression $\frac{d}{dx}(x^2)=2x$ can be viewed as a shorthand for

Let $f$ be the function defined by $f(x)=x^2$ for all $x$. The function $f'$, evaluated at the arbitrary point $x$, is equal to $2x$, i.e. $f'(x)=2x.$

Here, $x$ is simply a dummy variable used to illustrate what the derivative of the function is. It makes no difference whether we use $x$ or $y$ or any other variable. So $x$ is being used in two different ways here, hence why contradictory results appear.

By contrast, the second example, $x^2=9+y^2$, does show a relationship between the variables $x$ and $y$. Again, this could be expressed more formally as

Let $y$ be an arbitrary function satisfying $y(x)^2+9=x^2$ for all $x$.

When we differentiate both sides with respect to $x$, what we are doing is showing that $y'(x)$ must satisfy the following property:

$2x=2y(x)y'(x) \, .$

Even though we might not have an explicit formula about $y$, implicit differentiation allows us to make meaningful statements about this function.

Answer 3

A few notes: you are taking derivatives. I think the more fundamental operator is the differential (which gets confusing because taking a derivative is referred to as "differentiating"). The differential is like a derivative, but not with respect to any variable. So, for instance the derivative of $x^2$ is $2x$ but the differential is $2x\,dx$. The derivative of $xy$ depends on which variable the derivative is being taken with respect to, but the differential is just $x\,dy + y\,dx$.

Derivatives are more likely to lead to contradictions. For instance, take the equation $x = 1$. The derivative of this with respect to $x$ is $1 = 0$. However, the differential is $dx = 0$, which is true: the value of $x$ is not changing. The reason for the contradiction of the derivative is that the derivative with respect to $x$ is just the differential divided by $dx$. Since $dx = 0$, dividing both sides by $0$ leads to an invalid $\frac{0}{0}$.

So, going back to your problem, when you had $x^2 = 9$, the differential gives you a better picture. $2x\,dx = 0$. Why? Because at $x^2 = 9$, $x$ is not moving, so the differential is zero. If you try to divide by $dx$, you will wind up with an invalid $\frac{0}{0}$.

So, in your second case, differentiating gives you: $2x\,dx = 2y\,dy$. You can then solve for the derivative by algebraic manipulations.

So, in both cases, finding the differential works. In the first case, finding the derivative doesn't work because you wind up with a $\frac{0}{0}$ when you try.