It’s Professor Dave, let’s discuss implicit differentiation.
Thus far, we have been applying differentiation to functions where one variable is being expressed
in terms of another. In these, f of x is interchangeable with y, and some x terms will follow. But
there are plenty of other functions that aren’t set up exactly like this. For example, x squared
plus y squared equals nine is the equation of a circle with it’s center at the origin
and a radius of three. How would we take the derivative of this? In order to do it, we
need to use a technique called implicit differentiation, so let’s learn this now.
With the example we were just discussing, we know that we could solve for y in terms
of x, and then take the derivative as we already know how. But this can be difficult and time-consuming,
and sometimes it’s not even possible. Implicit differentiation will allow us to take the
derivative of this function without having to solve for one variable in terms of another.
Instead, we can just leave everything as it is, and simply differentiate with respect
to one specific variable. We typically differentiate with respect to x, so let’s do that to this
function, and then we can solve for y prime, or the derivative of y with respect to x,
which is what we are looking for. So the first thing we must do, as we said, is take the
derivative of both sides with respect to x. That is what this notation represents, with
d over dx next to these terms. Make sure to put each side in parentheses, because we need
this d in the numerator to operate on everything listed, as that is what this means, it means
the derivative of all of this, over dx, which means with respect to x. Once we do this,
it’s just a matter of rearranging things and evaluating. We know that the derivative
of a sum is just the sum of derivatives, so we can change this to the derivative of x
squared with respect to x plus the derivative of y squared with respect to x. On the right,
the derivative of any constant is zero, so that becomes zero. Now let’s evaluate these
terms. This is the part we have to really wrap our heads around. We are specifically
treating y as a function operating on x, and we should therefore imagine f of x anywhere
we see y. So when we take the derivative of x squared with respect to x, it’s just what
we are used to. Two comes down here, the exponent becomes one, and we get two x. But for y squared
it’s different, because we are not taking the derivative with respect to y. It is still
with respect to x, and again, we are thinking of y as a function that depends on x. For
that reason, we have to use the chain rule. Remember that the derivative of a composite
function is the derivative of the outer function times the derivative of the inner function,
and we imagine this as f of x raised to the second power. So here we will take the derivative
of y squared with respect to y, thinking of this exponent as the outer function operating
on y, but then also imagining y as some other expression with x’s in it, so we multiply
by the derivative of y with respect to x, which is dy over dx. The derivative of y squared
with respect to y is easy, that’s just two y as we would expect. Then we are left with
dy over dx, which we can also just call y prime, and this is precisely what we are looking
for, so let’s solve for it. Subtract two x from both sides, divide both sides by two
y, and we are left with negative x over y. So this is y prime, or the derivative of the
original function with respect to x. Here’s a slightly trickier one. X cubed
plus (x squared, y) plus four y squared equals six. Again, we will take the derivative of
everything with respect to x, and we can treat each term in the sum separately. X cubed is
easy since there’s no y. We just get three x squared as we would expect. Now for this
second term, we have to use the product rule, since it’s a product. So let’s write x
squared times the derivative of y plus the derivative of x squared times y. Now we evaluate
these derivatives. The derivative of y will be y prime. Remember, y is a function, and
its derivative is some other function, which is what we are looking for, so we just write
y prime. The derivative of x squared is two x. So we have x squared y prime plus two xy.
And then we move on to the four y squared term. Here we need to use the chain rule,
so we take the derivative with respect to y first, ignoring what’s inside y, giving
us eight y, but then we multiply by the derivative of y with respect to x, giving us y prime,
so that will be eight y, y prime. And six becomes zero. Remember, we do this extra stuff
with y only because we are differentiating with respect to x, and not y. Now let’s
bring all the terms without y prime in them to the other side. We do this so that we can
then factor y prime out of these two terms. We do this so that we can then divide both
sides by this parenthetical term, allowing us to solve for y prime, which is the point
of all this. Then let’s just factor a negative x out of the numerator for good measure, and
we are left with y prime equals negative x times the quantity three x plus two y, divided
by x squared plus eight y. Let’s try another example of similar difficulty
just to make sure we have this. How about (x squared, y) plus (x, y squared) equals
three x, and again we want to differentiate with respect to x. On the left we have two
products, so we have to use the product rule for each of them. Taking the first one, we
get x squared times the derivative of y, which is y prime. To that we add the derivative
of x squared, or two x, times y. Now we do the other product. This will give us x times
the derivative of y squared, and once again, any time we take the derivative of the variable
representing the function, and there are other operations on that variable, we must use the
chain rule. So the derivative of y squared will be two y times y prime. Again, this makes
sense if we imagine y as being some function full of x’s. If that were inside parentheses
and then squared, we already knew from learning the chain rule that the derivative of this
would be two times what’s inside the parentheses times the derivative of what’s in parentheses.
That’s what we are doing here, just with y representing whatever might be in there.
And then to finish the product rule, we get the derivative of x, which is one, times y
squared. On the other side, three x becomes three. Once again, let’s keep any terms
with y prime on this side and bring the other terms to the other side, so that we can factor
a y prime out of this sum. Once we have done that, we can then divide both sides by this
whole parenthetical term, which allows us to solve for y prime. That gives us three
minus two xy minus y squared all over x squared plus two xy.
Let’s do just one more that looks a little different. How about four cosine x sine y
equals one. Well first let’s divide both sides by four. That will make things a little
easier since one fourth will become zero anyway, and then we don’t have to deal with the
four. Now for the left side, we have a product, so let’s use the product rule. That gives
us cosine x times the derivative of sine y plus the derivative of cosine x times sine
y. Now for the derivatives, the derivative of sine y will require the chain rule. The
derivative of the outer function, sine, is cosine, so we get cosine y, but then we multiply
by the derivative of the inner function, y, which is y prime. Then for the other derivative,
cosine x will become negative sine x. Now we simply bring negative sine x sine y to
the other side, divide by cosine x cosine y, and we have solved for y prime. Just one
last step, we know that sine over cosine is tangent, so sine x over cosine x becomes tangent
x, and sine y over cosine y becomes tangent y, so y prime equals tangent x tangent y.
So that’s all there is to it. Implicit differentiation seems tricky at first, but once you get the
algorithm down, it’s no sweat, as we are just applying things that we already know,
like the product rule and chain rule. The only trick is to remember that if solving
for y prime, it means that y is some function in terms of x, and anytime we take the derivative
of y, if there are other things operating on y, we will have to use the chain rule,
which will produce the y prime terms that we end up solving for. Let’s check comprehension.