## Saturday, July 31, 2010

### Implicit Differentiation

I recently taught my summer Calculus 1 class about implicit differentiation.  This is usually a difficult subject to teach, as students have trouble understanding why they much include $\frac{dy}{dx}$ whenever they encounter a $y$ term.  I think I have stumbled upon a good way to explain this.

Before saying anything about how to differentiate, I take a few minutes to explain what an implicit function is.  In addition to showing some examples so students see the form of the equations, I point out that even thought the $x$'s and the $y$'s are "all mixed up," $y$ is still a function of $x$.  We just don't know what it is (because it is hard or even impossible to solve for $y$).

Next, I go through an example to see how to differentiate and then solve for $\frac{dy}{dx}$.  When I get to the $y^2$ term, I say that the derivative is $2y$, but because $y$ is a function of $x$, we need to use the chain rule, so we must also multiply by the derivative of $y$, which is $\frac{dy}{dx}$.  I point out that this is a good thing, since we are searching for the derivative of $y$, so one had better show up in our problem.  This, however, is not enough to convince most students.

Hopefully a student asks why we must include the derivative of $y$ (otherwise I ask them to explain in, which usually gets them to ask).  I remind them that $y$ is in fact a function of $x$.  I suggest that maybe $y = \sin(x)$.  Then $y^2 = \sin(x)^2$.  How do you take the derivative of that?  Chain rule.  Okay, what if $y = x^2 + 3$?  Chain rule again.  When writing these two examples up, I will not write out the derivative of the inside, instead leave it is $\sin(x)^\prime$ and $(x^2 + 3)^\prime$.  Then I go back and erase all the "$\sin(x)$" and "$x^2 + 3$" and replace the with $y$'s.  This seems to give students that elusive "ah-ha" moment.  And at least for the next day, they know how to do implicit differentiation.