11.7. Implicit Differentiation. Introduction. Prerequisites. Learning Outcomes

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1 Implicit Differentiation 11.7 Introuction This Section introuces implicit ifferentiation which is use to ifferentiate functions expresse in implicit form (where the variables are foun together). Examples are x 3 + x + 2 = 1, an x 2 a + 2 = 1 which represents an ellipse. 2 b2 Prerequisites Before starting this Section ou shoul... Learning Outcomes On completion ou shoul be able to... be able to ifferentiate stanar functions be competent in using the chain rule ifferentiate functions expresse implicitl HELM (2006): Section 11.7: Implicit Differentiation 51

2 1. Implicit an explicit functions Equations such as = x 2, = 1, = sin x are sai to efine explicitl as a function of x because x the variable appears alone on one sie of the equation. The equation x = x is not of the form = f(x) but can be put into this form b simple algebra. Task Write as the subject of x = x Answer We have (x + 1) = x 1 so = x 1 x + 1 We sa that is efine implicitl as a function of x b means of x+ +1 = x, the actual function being given explicitl as = x 1 x + 1 We note than an equation relating x an can implicitl efine more than one function of x. For example, if we solve x = 1 we obtain = ± 1 x 2 so x = 1 efines implicitl two functions f 1 (x) = 1 x 2 f 2 (x) = 1 x 2 52 HELM (2006): Workbook 11: Differentiation

3 Task Sketch the graphs of f 1 (x) = 1 x 2 f 2 (x) = 1 x 2 (The equation x = 1 shoul give ou the clue.) Answer Since x = 1 is the well-known equation of the circle with centre at the origin an raius 1, it follows that the graphs of f 1 (x) an f 2 (x) are the upper an lower halves of this circle. f = + 1 x 2 1 (x) 1 1 x 1 1 x f = 1 x 2 2 (x) Sometimes it is ifficult or even impossible to solve an equation in x an to obtain explicitl in terms of x. Examples where explicit expressions for cannot be obtaine are sin(x) = x 2 + sin = 2 2. Differentiation of implicit functions Fortunatel it is not necessar to obtain in terms of x in orer to ifferentiate a function efine implicitl. Consier the simple equation x = 1 Here it is clearl possible to obtain as the subject of this equation an hence obtain. HELM (2006): Section 11.7: Implicit Differentiation 53

4 Task Express explicitl in terms of x an fin for the case x = 1. Answer We have immeiatel = 1 x so = 1 x 2 We now show an alternative wa of obtaining which oes not involve writing explicitl in terms of x at the outset. We simpl treat as an (unspecifie) function of x. Hence if x = 1 we obtain (x) = (1). The right-han sie ifferentiates to zero as 1 is a constant. On the left-han sie we must use the prouct rule of ifferentiation: (x) = x + = x + Hence x = 1 becomes, after ifferentiation, x + = 0 or = x In this case we can of course substitute = 1 x to obtain = 1 x 2 as before. The metho use here is calle implicit ifferentiation an, apart from the final step, it can be applie even if cannot be expresse explicitl in terms of x. Inee, on occasions, it is easier to ifferentiate implicitl even if an explicit expression is possible. 54 HELM (2006): Workbook 11: Differentiation

5 Example 15 Obtain the erivative where x 2 + = Solution We begin b ifferentiating the left-han sie of the equation with respect to x to get: (x2 + ) = 2x +. We now ifferentiate the right-han sie of with respect to x. Using the chain (or function of a function) rule to eal with the 3 term: (1 + 3 ) = (1) + (3 ) = Now b equating the left-han sie an right-han sie erivatives, we have: 2x + = 32 We can make the subject of this equation: 32 = 2x which gives = 2x We note that has to be expresse in terms of both x an. This is quite usual if cannot be obtaine explicitl in terms of x. Now tr this Task requiring implicit ifferentiation. Task Fin if 2 = x2 + sin Note that our answer will be in terms of both an x. Answer We have, on ifferentiating both sies of the equation with respect to x an using the chain rule on the sin term: (2) = (x2 ) + (sin ) i.e. 2 = 2x + cos leaing to = 2x 2 cos. HELM (2006): Section 11.7: Implicit Differentiation 55

6 We sometimes nee to obtain the secon erivative 2 for a function efine implicitl. 2 Example 16 Obtain an 2 at the point (4, 2) on the curve efine b the equation 2 x 2 x 2 2 = 0 Solution Firstl we obtain We have from which b ifferentiating the equation implicitl an then evaluate it at (4, 2). 2x x 2 2 = 0 (1) = 2x x (2) so at (4, 2) = 6 10 = 3 5. To obtain the secon erivative 2 it is easier to use (1) than (2) because the latter is a quotient. 2 We simplif (1) first: 2x (x ) = 0 (3) We will have to use the prouct rule to ifferentiate the thir term here. Hence ifferentiating (3) with respect to x: or 2 (x )2 ( ) = ( ) 2 (x ) 2 Note carefull that the thir term here, be confuse with the secon erivative enote b 2 2. Finall, at (4, 2) where = 3 5 from which 2 = = 0 (4) 2 ( ) 2, is the square of the first erivative. It shoul not we obtain from (4): 2 at (4, 2). 2(3 5 ) 2( 9 25 ) ( )2 2 = 0 56 HELM (2006): Workbook 11: Differentiation

7 Task This Task involves fining a formula for the curvature of a bent beam. When a horizontal beam is acte on b forces which ben it, then each small segment of the beam will be slightl curve an can be regare as an arc of a circle. The raius R of that circle is calle the raius of curvature of the beam at the point concerne. If the shape of the beam is escribe b an equation of the form = f(x) then there is a formula for the raius of curvature R which involves onl the first an secon erivatives an 2. 2 Fin that equation as follows. Start with the equation of a circle in the simple implicit form x = R 2 an perform implicit ifferentiation twice. Now use the result of the first implicit ifferentiation to fin a simple expression for the quantit 1+(/) 2 in terms of R an ; this can then be use to simplif the result of the secon ifferentiation, an will lea to a formula for 1 (calle the curvature) in terms of R an 2. 2 HELM (2006): Section 11.7: Implicit Differentiation 57

8 Answer Differentiating: 2x + 2 = 0 x = R 2 gives: Differentiating again: From (1) = x 1 + ( ) 2 ( ) 2 R So 1 + =. Thus (2) becomes 2 so 2 2 = 1 R ( ) 3 R ( ) = 0 (2) 2 ( ) 2 = 1 + x2 = 2 + x 2 = 2 2 ( ) 2 R + 2 ( ) 2 = 0 2 (1) ( ) 2 R (3) 2 = 2 R2 3 ( ) ( ) 3 1 R = R Rearranging (4) to make 1 ( ) R R the subject an substituting for from (3) gives the result: 2 1 R = [ 2 ( ) ] 2 3/2 1 + The equation usuall foun in textbooks omits the minus sign but the sign inicates whether the circle is above or below the curve, as ou will see b sketching a few examples. When the graient is small (as for a slightl eflecte horizontal beam), i.e. is small, the enominator in the equation for (1/R) is close to 1, an so the secon erivative alone is often use to estimate the raius of curvature in the theor of bening beams. (4) 58 HELM (2006): Workbook 11: Differentiation