Math Implicit Differentiation. We have discovered (and proved) formulas for finding derivatives of functions like

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1 Math Implicit Differentiation Name We have iscovere (an prove) formulas for fining erivatives of functions like f x x 3x 4x. 3 This amounts to fining y for 3 y x 3x 4x. Notice that in this case, y is expresse explicitly in terms of x. But we nee not have y explicitly in terms of x to fin y. 1. Consier the relation x y 5. Graph the relation at right. (Note that this relation implicitly efines two functions.). Fin the slopes of the lines tangent to the graph where x 4 using the fact that any line tangent to a circle is perpenicular to the raius from the center to the point of tangency. Slope1 = Slope = 3. Now let's fin these slopes algebraically by using implicit ifferentiation. Notice that the relationship between x an y is efine implicitly within the equation. We say this because y is not alone on one sie of the equation. Start with the original equation, an ifferentiate both sies of the equation with respect to x. Since y is a function of x, be sure to apply the Chain Rule (an let y enote the erivative of y with respect to x). Solve for y in terms of x an y. Finally, use that fact that (4, 3) an (4, -3) are the points of tangency. 4. Extension of the Power Rule to Rational Exponents. p Slope1 = Slope = Let y x q. We might be tempte to use the power rule to fin y irectly here, but so far, we have only prove that the Power Rule works for integer powers, not for rational powers in general. Let's use implicit ifferentiation to see if the Power Rule works for rational powers. Raise both sies to the q power. Then we have integer exponents, so the Power Rule applies. Now ifferentiate implicitly an solve for y.

2 5. The figure at right shows the graph of the relation x 4xy 4y 10x 5. Verify that the points (3, 4) an (3, -1) lie on the graph. 6. Fin the equations of the lines tangent to the curve at (3, 4) an (3, -1). [Hint: Use implicit ifferentiation.] Then graph the lines above to make sure they are reasonable. Line 1: Line : 7. Now see if you can solve the equation of the relation given in Problem 5 for y in terms of x. [Hint: Use the Quaratic Formula.] 8. Which woul be easier Using implicit ifferentiation, as we i here, or to solve for y explicitly an then ifferentiate?

3 Derivatives of Inverse Trigonometric Functions 1 Recall that the function y sin x is the angle whose sine is x. We wish to erive (i.e., come up with) a formula for the erivative of this function Consier the graph of y sin x at right. What is the range of the function?. Take the sine of both sies of the equation, an raw a triangle that shows the relationship between x an y. 3. Now ifferentiate implicitly an solve for y in terms of x. Hint: Sincecos ysin y 1, we know that cos y But your answer to (1) implies that cos y 0. Thus we choose the positive branch. That is, 1 sin y. cos y 1 sin y Derivative of Inverse Sine Function 1 sin x sin 1 u 4. Use the formula you just erive to fin f x if f x sin 1 3x Fin g x if g x cos sin x. The function y 1 sec x is the angle whose secant is x.

4 1 6. Consier the graph of y sec x at right. What is the range of the function? 7. Take the secant of both sies of the equation an ifferentiate implicitly with respect to x. For now, just solve for y in terms of y. 8. Now we will express y in terms of x. Consier the trigonometric ientity Diviing both sies by an thus cos x gives the ientity tan y sec y 1 = 1tan y sec y, x 1 (two branches) If x > 0, then 0 y, which means the angle y is in the first quarant. Thus tan y is (positive or negative) So we choose this branch for tan y, an we have tan y = Thus, since x sec y, 1 sec ytan y If x 0, then y, which means the angle y is in the secon quarant. Thus tan y is (positive or negative) So we choose this branch for tan y, an we have tan y = Thus, since x sec y, 1 sec ytan y cos y sin y 1. So we have Derivative of Inverse Secant Function 1 sec x sec 1 u 9. Does this result agree with the graph shown above (i.e., the erivative is always positive)?

5 By similar methos, we can prove that cos 1 u u 1 u, tan 1 u u, 1 u cot 1 u u u 1 an 1 u csc u u u At a prison facility in Colorao, a camera is locate 100 ft from the wall of the prison, an has an infrare sensor that causes the camera to aim at anything that moves along the wall. a. Express the angle in terms of x. b. Suppose a prisoner is running along the wall (away from the guar tower) at a spee of 15 ft/s. How fast will the camera have to rotate on its mount in orer to keep the prisoner in focus at the instant that the prisoner is 10 ft from the tower? What if the prisoner is irectly uner the tower? Report your answers in egrees per secon =