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Introduction to Implicit Differentiation
Q: What is the difference between f(x) = 2x3 + 4x – 1 and y = 2x3 + 4x – 1? A: Essentially none. If you really want to point out a difference, it is the use of the notation: f(x) vs. y. If the notation f(x) is used, f(x) = 2x3 + 4x – 1 is of course a function. If y is used instead f(x), despite you can say y is a function of x, y = 2x3 + 4x – 1 is usually called a(n) equation. There are many kinds of equations, but mostly they fall into these two categories: explicit equations and implicit equations: Examples of explicit equations: y = 2x3 + 4x – 1, y = (2x – 1)/(4 sin x), y = sin x cos x Examples of implicit equations: 4x2 + 2y = y + 3, x2 + y2 = 25, y = cos y + x sin y Q: What is the distinction between an explicit equation and an implicit equation? A: ______________________________________________________________________________________________ Find dy/dx (i.e., y ) of the six equations above: 1. y = 2x3 + 4x – y = (2x – 1)/(4 sin x) 3. y = sin x cos x dy/dx = 6x dy/dx = cos x cos x + sin x(–sin x) 4. 4x2 + 2y = y x2 + y2 = y = cos y + x sin y y = –4x y2 = 25 – x (No way y can be easily written in terms dy/dx = –8x of x. So we can’t differentiate the equation explicitly, but we can differentiate it implicitly, see next page)

What is the Derivative of y with Respect to x?
Q: What is the derivative of x with respect to x? A: Q: What is the derivative of y with respect to x? Q: What is the derivative of y2, y3, 5y4, sin y, cos y, tan y, etc with respect to x? A: See below d/dx[(2x + 1)2] = d/dx[(2x + 1)3] = d/dx[5(2x + 1)4] = d/dx[y2] = d/dx[y3] = d/dx[5y4] = d/dx[sin (2x + 1)] = d/dx[cos (2x + 1)] = d/dx[tan (2x + 1)] = d/dx[sin y] = d/dx[cos y] = d/dx[tan y] = Find dy/dx (i.e., y ): 1. 4x2 + 2y = y x2 + y2 = y = cos y + x sin y Compare this with the last page, the denominator y is exactly

Implicit Differentiation (i.e., Finding dy/dx in an Equation)
Example 1: x2 + y2 = 25 Q: What can we do with the dy/dx? A: We can use dy/dx to find the slope of the tangent line at a given point and hence the equation of the tangent line at that point. Find the equation of the tangent lines at (3, 4) and (–4, –3). How to find dy/dx (i.e., y) of an equation? Just differentiate term by term with respect to x: If a term only involves with x, just differentiate regularly. If a term only involves with y, differentiate with respect to y and then multiplied by dy/dx, or y (we are actually using the chain rule here). If a term involves a product (or quotient) of x and y, use product rule (or quotient rule). If the term has a power, e.g., (expression)n, d/dx(expression)n = n(expression)n–1· d/dx(expression). Example 2: x + 2y = xy Example 3: y = cos y + x sin y Isolate dy/dx (i.e., the y) by: Get the dy/dx terms on one side and the terms without dy/dx on the other. Factor out the side with the dy/dx terms. Divide both sides by the coefficient of dy/dx.

Implicit Differentiation (cont’d)
Example 4: Find the equation of the tangent line at (1, 1) of the equation (x2 + y)2 = x + 3y Find the equation of the tangent line at (0, 0) and at (1, 1) of the equation above. How to find dy/dx (i.e., y) of an equation? Just differentiate term by term with respect to x: If a term only involves with x, just differentiate regularly. If a term only involves with y, differentiate with respect to y, then multiplied by dy/dx, or y (we are actually using the chain rule here). If a term involves a product (or quotient) of x and y, use product rule (or quotient rule). If the term has a power, e.g., xpression)n– (expression)n, d/dx(expression)n = n(e1· d/dx(expression). Isolate dy/dx (i.e., the y) by: Get the dy/dx terms on one side and the terms without dy/dx on the other. Factor out the side with the dy/dx terms. Divide both sides by the coefficient of dy/dx. We still need to isolate dy/dx, however, we don’t need to do the algebra in terms of x and y if we have numbers to substitute for x and y. Just plug the numbers into x and y first and then isolate the dy/dx.

Why Do We Differentiate with Respect to x and not to y?
As we have seen earlier, when we are given an equation and are asked to find dy/dx, it really means we need to differentiate the equation with respect to x. So do we ever and can we differentiate the equation with respect to y too? Of course we can. However, we don’t usually differentiate with respect to y (i.e., under normal circumstances, we won’t be asked to find dx/dy). This is why: the notation dy/dx really comes from the familiar notation _____, which is used to find the slope of a line, and of course, we use dy/dx to find the slope of the _______ line. Since we never use x/y to find the slope of anything, therefore, we almost never will be asked to find dx/dy. If we are asked to find dx/dy, the way to find it is similar (parallel) to finding dy/dx. This is how: Find dy/dx and dx/dy of the following equation: 1. x2 + y2 = x2 + y2 = 100 Whereas the above equation on the left is differentiated with respect to x, the (same) equation on the right is differentiated with respect to y. And we actually can differentiate an equation with respect to anything we want: 1. Differentiate the equation with respect to z: x2 + y2 = 100 2. Differentiate the equation with respect to t: x2 + y2 = 100 Recall: When differentiating a term in a variable, say a, with respect to another variable, say b, use Chain Rule: differentiate the term with respect to a first, then to b.

Differentiate with Respect to Time
If we have an equation in terms of x and y, and if we are going to differentiate the equation, but we don’t differentiate it with respect (wrt) to either x or y, we will most definitely differentiate it wrt __, i.e., wrt _____. Problem 1: A ladder 10 ft long is placed against a wall. The top of the ladder touching the wall is slipping downward ½ ft per min, how fast is the bottom of the ladder pushing away from the wall when the top of the ladder is 8ft above ground? The problem above is called a related rates problem. A related rates problem is a problem that deals with the rates of change of two (or more) variables. The rates of change of one thing, a, wrt another, b, is the _________, i.e., ______. Most relate rates problems deal with rates of change wrt _____, therefore it’s almost always d/dt, i.e., the derivative of something wrt time. In the problem above, the two rates of changes are _____—how fast the top of the ladder is slipping down and dx/dt—how fast ___________________________. More related rates problems are given below and next page: Problem 2: The area of a circle is expanding 10 sq in per min. How fast is the radius increasing when the radius is 6 in? Problem 3: The radius of a sphere is expanding 2 cm in per sec. How fast is the volume of sphere expanding when the radius is 8 cm?

More Related Rates Problems
Problem 4: A water tank has the shape of a inverted cone with base radius 2 m and height 4 m. Water is being poured into the cone. If the radius is changing ½ m per min and the height is changing 1 m per min, what is the rate of change of the volume at the instant when the radius is 1 m and height is 2 m? We might never find problems like problem 4 in the textbook because it involves three rates of change—radius, height and volume. All problems in the textbook deals with at most two rates of change. In the text book, this problem will be stated as such: How to solve a related rate problem: Read the problem carefully. Draw a diagram if necessary. Determine what you are given and what you need to find. Find an equation or a formula that relates the variables. Differentiate the equation with respect to time, appropriately use implicit differentiation and chain rules. Substitute the known values of the variables and rates and solve for what you need to find. Problem 5: A water tank has the shape of a inverted cone with base radius 2 m and height 4 m. Water is being poured into the cone. If the radius is changing ½ m per min, what is the rate of change of the volume at the instant when the radius is 1 m? Problem 6: At noon, ship A is 100 km west of ship B. Ship A is sailing east at 30 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 3:00 PM?

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