MA 2231 Lecture 18 - Higher Order , Implicit Differentiation

Friday, March 15, 2019

dy Objectives: Introduce dx notation, higher order derivatives, and implicit differentiation.

Derivative Notation

For a f, we have been using a prime, as in f 0 to indicate the function. According to Wikipedia, this notation is named after Joseph-Louis Lagrange, one of the pioneers in the development of . We generally think of Isaac Newton and Gottfried Leibniz as being the inventors of calculus, but many of the basic ideas go back much further.

Leibniz’s Notation. The derivative is a slope, change in y (1) . change in x We find the derivative by taking the slope of a secant line through two points on the graph of f, and then take the limit as one point approaches the other, and both the change in y and change in x go to zero. If you think of these limits as infinitesimals (infinitely small numbers), dy and dx, the derivative is the fraction

0 dy (2) f (x) = . dx It’s useful sometimes to think of derivatives this way, although it’s best to not take it too far. In any case, the notation is commonly used, and for a function (3) y = f(x), all of the following represent the derivative of f:

dy 0 df d 0 (4) = f (x) = = f(x) = y . dx dx dx

Newton’s and Euler’s Notations. These might come up, but only in passing. Newton’s Notation indicates a derivative with a dot over the letter, and usually is associated with an input variable representing time, so the derivative might be written dy (5) =y. ˙ dt The dot notation seems to be most common in physics and engineering, but I’ve seen it in biomechanics. For example, endurance athletes are interested in their VO˙ 2max. The V is for volume of oxygen, and V˙ is the volume of oxygen per minute (the derivative of V with respect to the time variable t). In Euler’s Notation, the derivative is indicated by a D, so 0 (6) f (x) = Df(x).

Higher Order Derivatives

We have been taking the derivatives of functions, and the result is another function. We can, if we wish, take the derivative of the derivative. And we can keep taking derivatives as many times as we want. We’ve seen one reason to do this already in the lab yesterday. If the derivative of f 0 is positive, then f 0 is increasing, and this means that f must be curving up (we’ll eventually call this concave up). 1 MA 2231 Lecture 18 - Higher Order Derivatives, Implicit Differentiation 2

The . The derivative of the derivative of f will be called the second derivative of f. For y = f(x), the second derivatives will be written 2 2 00 d y d 00 (7) f (x) = = f(x) = y =y ¨ = D2f(x) = etc. dx2 dx2

Third Derivatives and Higher. The third derivative will be written in the obvious way 3 000 d (8) f (x) = , dx3 but the primes start to pile up, so beyond the third derivatives, we will use a sort of Roman numeral sort of notation, and then a number in parantheses like (9) f iv = f (4), f v = f (5), etc. and in general, f (n) for the nth derivative.

Quiz 18A

1. Find the first, second, third, and fourth derivatives of y = f(x)=3x3 4x2 + x 17. − − d10y 2. What is dx10 ?

Implicit Differentiation

Most of the time, we have a function y = f(x), like (10) y = f(x) = x3 +7, and we’ll say that “y is a function of x.” The equation (11) y = x3 +7 explicitly defines y in terms of x. If we were to solve this equation for x, (12) x = p3 y 7, − we can say that this equation “explicitly defines x as a function of y.”

dy dx There are, therefore, two derivatives that we can talk about, dx (the derivative of y with respect to x) and dy (the derivative of x with respect to y). These two derivatives are very closely related, but they’re definitely not the same. Basic Principle 1. When we work with a function, and take a derivative of that function, we are choosing which variable is the input variable and which is the output variable. There is no correct way, and someone is simply choosing one. We’ll typically think of y as being a function of x.

Now, consider the equation (13) x2 + y2 =4. We can think of one of these variables as being a function of the other, as usual, we’ll think of y as being a function of x. We’ll say that this equation implicitly defines y as a function of x (and vice versa). This “implicit” equation is in some ways preferable to the “explicit,” because we have to split it into two explicit equations (14) y = p4 x2. ± − These are the top and bottom halves of the circle of radius 2. MA 2231 Lecture 18 - Higher Order Derivatives, Implicit Differentiation 3

So how do we take the derivative with an implicit equation? We simply have to be very careful about what’s a function of what. In this case, we’re thinking of y as being function of x, so the derivative of y with respect to x is 0 d dy 0 (15) [ y ] = [ y ] = = y = etc. dx dx Next, we’re going to have to differentiate y2. Since y is a function of x, we need to use the 0 d dy (16)  y2  =  y2  =2y . dx · dx Now, x can be thought of as being a function of x, but dx d 0 (17) = [ x ]=[ x ] =1, dx dx so whether we use the chain rule or not, it doesn’t matter. In either case, d (18)  x2  =2x. dx

Taking a derivative implicitly. We have this implicit equation (19) x2 + y2 =4, and to differentiate implicitly with respect to x, we take the derivative of both sides of the equation thinking of y as a function of x. We get dy (20) 2x +2y =0. · dx Basic Principle 2. In implicit differentiation with respect to x, differentiate both sides of the equation treating x as you normally would and y as a function of x with the chain rule.

We can find the derivative explicitly by differentiating (the two functions) 1 (21) y = p4 x2 = (4 x2) 2 . ± − ± − We get

dy 1 − 1 x (22) = (4 x2) 2 2x = − . dx ±2 − · − ±√4 x2 − Note that this result agrees with what we got implicitly (substituting y = √4 x2). ± − Interpretation of an implicit differentiation. In the equation dy (23) 2x +2y =0, · dx we are saying that for any point (x, y) on the graph (in this case the circle of radius 2), x, y, and the slope dy √ √ of the tangent line m = dx satisfy this equation. For example, at the point ( 2, 2), the tangent line has slope dy = 1. Plugging these in, we get dx − (24) 2(√2)+2(√2) ( 1)=0, · − we see that these values satisfy the equation. We can also find the slope of the tangent line. For example, at (√3, 1), we have the equation dy (25) 2(√3) + 2(1) =0. dx dy Solving the equation for dx , we get dy 2√3 (26) = = √3. dx 2 MA 2231 Lecture 18 - Higher Order Derivatives, Implicit Differentiation 4

Examples

Example 1. Differentiate the equation (27) y3 3y2 + y 14=3x4 − − implicitly with respect to x. We get dy dy dy (28) 3y2 6y + = 12x3. dx − dx dx Example 2. Differentiate the equation (29) x2y3 = x + y implicitly with respect to x. Don’t forget the . We get dy dy (30) x2 3y2 + y3 2x = 1 + . · · dx · dx

Quiz 18

Differentiate the following implicitly with respect to x.

1. x3 y3 = 4. − 2. xy = 7.

Homework 18

Differentiate the following implicitly with respect to x.

1. 3y7 + x = 4.

2. y2 + y + 1 = ln(x).

3. x2y2 = 1.

4. sin(y) + sin(x)=1.

5. ln(y) = x.

6 dy Answers: 1) 21y dx +1=0. dy dy 1 2) 2y dx + dx = x . 3) x2 2y dy + y2 2x = 0. · dx · dy 4) cos(y) dx + cos(x)=0. 1 5) dy = 1. y dx