, Sometimes you don't have a function, you have a relation between two variables. A
classic example is x squared plus y squared equals, say, 25. The graph of the
points in the plane that satisfy this equation as a circle. But that's not the
graph of a function, right? This graph fails the vertical line test. For a given
input value, say 4 in this case, there's multiple y values which satisfy this
equation. So, I can't simply solve this equation for y. Nevertheless, if you pick
a specific point like 4, 3, you might be able to find a function whose graph traces
out that same curve. So yeah, if I pick 4, 3, there is a function, y equals the
square of 25 minus x squared. Which traces out a piece of the whole curve, right? I'm
just ignoring the rest of this and this little tiny piece of the curve can be
regarded as a function. If I had picked a different point, then I'm going to pick a
different function. Instead of the square root of 25 minus x squared, if I wanted to
stand down here, near the point 4, minus 3, well then maybe I'd pick the function y
equals negative the square root of 25 minus x squared. If I ignore the rest of
this and I'm just looking at this curve here, yet this curve by itself is a
function. If I ignore this, it satisfies the vertical line test. This function is
picking out a piece of the curve given by this equation which is, yeah, only valid
near the point 4, minus 3. But maybe that's all I care about for the time
being. So, let's say there is a function, y equals f of x, that satisfies the
original equation. Well then, I can write that down. y equald f of x say satisfies
the equation just means that x squared plus f of x squared equals 25. Now, I'm
not saying that this gives me all of the solutions, right? The graph x squared plus
y squared equals 25 is a circle fails the vertical line test. There is no function
that gives me all those outputs because there's multiple outputs for a given
input. All I'm saying is that I've got some function which traces out a piece of
the whole curve. Then, I can differentiate. So, this is true for a
bunch of values of x that I can differentiate this. The derivative of this
sum is the sum of the derivative, so the derivative of x squared is 2x plus the
derivative of f of x squared. I'm going to use a chain rule to do that. It's the
derivative of the outside function at the inside times the derivative of the inside
function equals the derivative of 25, which is zero. Now I can solve. So,
subtract 2x from both sides and I'm left with 2 times f of x times f prime of x
equals negative 2x. And then, I'll divide both sides by 2 times f of x. And I'll
find that f prime of x is minus 2x over 2 f of x, and I can cancel those 2's and
just get minus x over f of x. It seems like a funny situation. The derivative
depends on more than just x. It also has an f of x. in it.
Another way to say it is that the slope of the tangent line dy, dx, is negative x
over y, right? y is f of x. And it does really seem a littie bit off putting
initially in these kinds of calculations the slope of the tangent line depends on
more than just x. It's negative x over y for this particular case. But think back
to the piacture for this case, right? The picture's a circle. And what I'm
saying is the slope of the tangent line is negative x over y. So, if you pick that
point, say 4,3, and you ask what's the slope of the tangent line to the circle at
the point 4,3 this equation is telling you the slope is -4 thirds. And yeah, that
line is going down, the slope's negative. What's the slope of the tangent line to
the curve at the point 4, negative 3? Same equation tells us that the slope there is
4 3rds. And yeah, this line's going up. The slope of the tangent line is depending
on more than just the x coordinate, right? You also need to know the y coordinate in
order to know exactly what function you're actually looking at near that point. And
that totally affects the slope of that tangent line. To do all these sorts of
calculations, the trick is the cha in rule.
For instance, if you're given some relation like this, x squared plus y cubed
equals 1. You just got to make sure to think of y as a function of x. So that
when you differentiate both sides, the derivative of the left hand side is 2x
plus the derivative of y cubed equals the derivative of 1, which is 0, but what's
the derivative of y cubed? If y is a function of x, then when you differentiate
this, you've got to use the chain rule. It's 3 times the inside function squared,
that's the derivative of the third power function, times the derivative of the
inside function. I'll just write y prime. And as long as you're careful to use the
chain rule, you'll be able to do these kinds of implicit differentiation
problems. And you'll eventually solve for y prime in terms of both x and y. The
chain rule is our friend.