1 Lecture 20: Implicit differentiation
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1 Lecture 20: Implicit ifferentiation. Outline The technique of implicit ifferentiation Tangent lines to a circle Derivatives of inverse functions by implicit ifferentiation Examples.2 Implicit ifferentiation Suppose we have two quantities or variables x an y that are relate by an equation such as x 2 + 2xy 2 + x 3 y = xy. If we know that y = y(x) is a ifferentiable function of x, then we can ifferentiate this equation using our rules an solve the result to fin y or y/x. In this course, we will not learn conitions which guarantee that y is a ifferentiable function of x. This is a topic for a later course. This assumption is usually vali an the technique is very useful. We begin with a simple example to practice the basic skill of ifferentiation as it is neee in implicit ifferentiation. Example. Differentiate the expression below with respect to x. Assume that y = y(x) is a function of x. x (x2 y 3 ) Differentiate the expression below with respect to y. Assume that x = x(y) is a function of y. y sin(x + y2 ). Solution. For the first one, we begin by noting that x y3 = 3y 2 y x Then using the prouct rule, x x2 y 3 = 2xy 3 + 3x 2 y 2 y x. For the secon one, we use the chain rule again to obtain y sin(x + y2 ) = ( x y + 2y) cos(x + y2 ). by the chain rule.
2 We show how to use this technique to fin tangent lines to a circle. Example. Consier the circle centere at the origin with raius 5 which is the set of points (x, y) which satisfy x 2 + y 2 = 25. Fin y/x on the circle. Fin the tangent lines at the points on the circle with x-coorinate 4. Show that a tangent line to the circle is perpenicular to the raius at the point of tangency. Solution. We imagine that y = y(x) is a function of x in the equation efining the circle an ifferentiate both sies with respect to x. x (x2 + y(x) 2 ) = x 25 2x + 2y y x = 0. Observe that when we ifferentiate the term y(x) 2, we use the chain rule with y(x) as the insie function. Next, we solve this equation for y/x to fin y x = x/y. To fin the tangent lines when the x-coorinate is 4, we solve 4 2 +y 2 = 25 for y to fin that y = 3 or 3. Thus there are two points where we nee to fin the tangent line. One passes through the point (x, y) = (4, 3) an has slope y/x = 4/3. The secon passes through (x, y) = (4, 3) an has slope y/x = 4/3. The point slope forms of the equation are: y 3 = 4 (x 4) 3 y + 3 = 4 (x 4) 3 The following sketch shows the tangent lines an the circle an helps to check our answer.
3 y x Next, at a general point (x, y) on the circle the tangent line has slope x/y while the raius which is the line segment joining (x, y) to (0, 0) has slope y/x. The prouct of these slopes is - an hence the lines are perpenicular. Exercise. We can also fin tangent lines by solving the equation x 2 + y 2 = 25 to give y = ± 25 x 2 an then using techniques we learne earlier. Carry this out to check your answer to the previous problem. Example. Fin the secon erivative y at the point (3, 4) on the circle x 2 +y 2 = 25. Note that in this problem we use the notation y for the erivative of y with respect to x, rather than the Leibniz notation, y/x. Solution. obtain We begin as before by ifferentiating x 2 + y 2 = 25 with respect to x an 2x + 2yy = 0. () As before, we have y = x/y an we woul like to ifferentiate again. It is probably simpler to ifferentiate () rather than y = x/y to avoi using the quotient rule. Differentiating both sies of () with respect to x an using the prouct rule on the secon term gives 2 + 2yy + 2(y ) 2 = 0.
4 Solving for y gives y = (x + (y ) 2 )/y = y x2 y 3. In the secon step we use that y = x/y. Now we may substitute the values (x, y) = (3, 4) to obtain y = /4 9/64 = 25/64. Example. Suppose that s an t are relate by the equation s 2 +te st = 2. Fin s/t. Solution. We assume that s is a function of t, s(t), ifferentiate both sies of the equation efining the curve an group the terms involving s/t obtaining t (s2 + te st ) = t 2 2s s t + est + t(s + t s t )est = 0 (2s + ste st ) s t + stest = 0 We use the prouct rule an the chain rule to carry out the ifferentiation. Solving for s/t gives s t = stest 2s + ste. st Note that even if the equation relating s an t, the equation for s/t is a linear equation an is easily solve..3 Derivatives of inverse functions The technique of implicit ifferentiation can also be use to fin the erivative of inverse functions. We illustrate this by fining the erivative of the function sin (x). Example. Solution. Fin the erivative of the inverse sine function sin or arcsin. If y = sin (x), then we have that sin(y) = x. Differentiating equation with respect to x an recalling that y = y(x) is a function of x gives that y cos(y) = or y = cos(y).
5 In orer to simplify this last expression, we recall the pythagorean ientity, sin 2 (y) + cos 2 (y) = or cos(y) = ± cos 2 (y). Our efinition of the sin tells us that y is in the range [ π/2, π/2] an thus that cos(y) 0. Thus we have cos(y) = sin 2 (y) = sin 2 (sin (x)) = x 2. This gives the expecte result that x sin (x) = x 2..4 Aitional examples Example. Fin the tangent line to the curve efine by x 2 + 2y 2 = 2 + x 2 y at the point (x, y) = (3, ). Solution. The tangent line will go through the given point (3, ) thus the only thing we nee to fin is the slope, y. We visualize that y = y(x) is a function of x an ifferentiate both sies of the equation (x 2 + 2y(x) 2 ) = (2 + x 2 y(x)) where enotes the erivative with respect to x. We use the prouct an chain rules to conclue 2x + 4yy = 0 + 2xy + x 2 y. We solve this equation for y an obtain Substituting (x, y) = (3, ) gives y (4y x 2 ) = 2xy 2x or y = 2xy 2x 4y x 2. y = = 0. Thus the tangent line is the line through (3, ) with slope 0 which gives y =. In our last example, we will not use x an y. It is useful to remember that the technique of implicit ifferentiation can be use to fin the rate of change between any two variables. Example. Consier the quaratic equation x 2 + 2x + c = 0. a) Fin the roots when c = 0. b) Fin the erivative of x with respect to c an for each root from part a) etermine if the root increases or ecreases as c increases. c) Sketch the parabola y = x 2 + x + c for c = 0 an check if your answer in part b) makes sense.
6 Solution. a) When c = 0, the equation x 2 + 2x = 0 factors as x(x + 2) = 0. The roots are x = 0 an x = 2. b) We ifferentiate the equation with respect to c an fin Solving for the erivative gives 2x x c + 2x c + = 0. x c = 2x + 2. At x = 0, we have x/c = /2 so this root ecreases as c increases. At x = 2, we have x/c = /2 so this root increases. c) As c increases, the parabola is shifte up an the roots move towars x =. c=0 c=/2 5 y x -.5 Exercises. Fin y/x when x an y are relate as follows: (a) y 2 + xy = 2 (b) e xy + xy + x 3 = xy 2 (c) sin(xy) + cos(xy) = /2
7 2. Fin x/z when x an z are relate as follows. (a) x 2 z 2 = (b) x 2 + axz + x sin(z) = 2 3. Consier the curve efine by x 2 + xy = 3. (a) Fin the value(s) of x when y = 2. (b) Fin all tangent lines to the curve at points (x, y) with y = Let y = tan (x) be the inverse tangent or arctangent function. Fin the erivative y/x by applying implicit ifferentiate to the equation x = tan(y). This provies another way to unerstan our metho for fining erivatives of inverse functions. October, 205
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