Lean Calculus

I’m thinking of a function with domain variable, say \(x\text{.}\) Let’s call it \(y\text{.}\) Suppose that I told you that \(x^2y=1\text{.}\) Notice that here I have not *explicitly* defined \(y\) in terms of \(x\text{.}\) Can you compute the derivative of \(y\text{?}\) Probably you realize that by dividing both sides by \(x^2\text{,}\) you find the *explicit* expression \(y=1/x^2\) so that you can then compute the derivative directly \(y'=-2/x^3\text{.}\)

But, sometimes equations can be very complicated and finding such explicit expressions for the function implied there just might not be possible. So, suppose we pretend for a moment that we can’t solve \(x^2y=1\) explicitly for \(y\text{.}\) I know, we can and we did just above. But, play along like we can’t. Is there nothing we can do to get information about \(y'\text{?}\) Well, there is. Since you know that \(y\) is *some* function of \(x\text{,}\) that means that \(x^2y\) is a product of two functions of \(x\) (namely \(x^2\) and \(y\)) We also know from previous sections how to differentiate a product. So, differentiation both sides of the equation \(x^2y=1\) gives \(2xy+x^2y'=0\text{.}\) Solving for \(y'\) gives \(y'=-2xy/x^2=-2y/x\text{.}\) This doesn’t give \(y'\) explicitly yet since we don’t know \(y\) explicitly yet. But, if someone came along and told us that \(y=1/x^2\text{,}\) notice that this conclusion gives us immediately that \(y'=-2/x^3\text{.}\)

Of course, it may never happen that you explicitly know what \(y\) is so that getting an explicit description of \(y'\) might not be possible. So, does this mean we have no information about \(y'\text{?}\) Quite the contrary. Suppose for example that due to some physical knowledge of the situation represented by the equation, we knew that the function \(y\) had the value \(1\) at \(x=1\) (i.e. \(y(1)=1\)). Then the equation we found, \(y'=-2y/x\text{,}\) tells us that \(y'(1)=-2(1)/1=-2\text{.}\) So, we know the slope of the tangent line to the graph of \(y\) at \((1,1)\) even though we have no explicit formula helping us to graph \(y\text{.}\) Now that we have the slope of the tangent line, \(2\text{,}\) and the point on the graph it passes through, \((1,1)\text{,}\) we can write an equation for the tangent line: