Clarifications about implicit differentiation

Question

Consider what happens when you take $$\frac{d}{dx}y^2$$

To me, this makes little sense as the expression $y^2$ does not even contain the variable $x$. So how can we differentiate with respect to $x$? Could we consider $y^2$ as $y^2x^0$? We could then treat the $y^2$ term as a constant, and then apply the power rule. However, this is clearly not the solution. What is wrong with the above argument?

The chain rule is required to work the above out. However, it is not clear why we even need the chain rule above. Consider what happens when finding the derivative $\frac{d}{du}$ of $uv$. How does it become $u\frac{dv}{du}+v$?

Could someone please give me some resources to help me understand implicit differentiation?

Answer 1

I'd recommend https://maa.org/press/maa-reviews/calculus-an-intuitive-and-physical-approach for a really great explanation. And wikipedia also has a good explanation (read the begining of the page a couple times) https://en.wikipedia.org/wiki/Chain_rule

But to you are right, the trick here is the chain rule.

To your first question about $\frac{d}{dx}y^2$. If we use the chain rule then we get $$ \frac{d}{dx}y^2 = \frac{d(y^2)}{dy} \frac{dy}{dx} = 2y\frac{dy}{dx} $$

The cool thing there is the last bit $\frac{dy}{dx}$. If $y$ indeed had no dependence on $x$, then that last bit would be zero (the derivate is the rate of change with respect to the variable you are deriving). But if $y = y(x)$ then that last factor is sort of like saying "I don't know if there is a dependence on $x$ but just in case there is".

If you were to consider $y^2$ as a contant, then $\frac{dy}{dx}$ would be zero. So you can, but you gotta remember that that last bit has to be "interpreted".

The last bit you add to your question is what makes thing work and it is very important. When you use the chain rule, you sort of introduce any variables that you think may affect your answer. For example, if $v$ were a function of $u$ ($v = v(u)$), and $u$ a function of $x$ ($u = u(x)$) then $$ \frac{d v}{dx} = \frac{dv(u)}{dx} = \frac{dv(u)}{du} \frac{du(x)}{dx} = \frac{dv}{du} \frac{du}{dx} $$

Answer 2

What you miss is that $y$ is treated as a function of $x.$

We have some equation of the form $F(x,y)=0.$ This equation describes a curve in the plane. For sections of this curve there exists some function $f$ such that the section is the graph $y=f(x),$ i.e. $F(x,f(x))=0$ for $x$ in some interval. What we do in implicit differentiation is that we identify $y$ with $f(x)$ so $y'=f'(x).$

An example might clarify. Take $F(x,y)=x^2+y^2-1.$ The curve is then the unit circle with center at origin. For this curve we can find an explicit formula for the upper part of this circle: $y=\sqrt{1-x^2}.$ The derivative of this is $y=-x/\sqrt{1-x^2}$ but let us take the derivative of both sides of $x^2+y^2-1=0$ where we treat $y$ as a function of $x$; this gives $2x+2yy'(x)=0,$ i.e. $y'(x)=-x/y.$ Inserting the expression for $y$ we again get $y'(x)=-x/\sqrt{1-x^2}$ as expected.