IMPLICIT DIFFERENTIATION CALCULUS 3 INU0115/515 (MATHS 2) Dr Arian Jannetta MIMA CMath FRAS Implicit Differentiation 1/ 11 Arian Jannetta
Explicit an implicit functions Explicit functions An explicit function is one where the link between the inepenent variable (say x) an the epenent variable (say y) is clearly efine. Here are some examples of explicit functions: y= x 2 + 2 y=3sin2x+2cos2x y= 5+xe x In each case, the value of y can be calculate from a value of x. So far, we have only carrie out ifferentiation of explicit functions. Implicit functions An implicit function is one where the relationship between the inepenent variable an epenent variable is not given in an explicit form. For example: x 2 + y 2 = 4 sin y= e xy xy 2 = 8x lny It is sometimes possible to rearrange an implicit function to obtain an explicit function. However, if we want to ifferentiate then we can o it using implicit ifferentiation. Implicit Differentiation 2/ 11 Arian Jannetta
Implicit functions Implicit functions have the general form F(x, y) = 0. The LHS is rea as function of x an y. Implicit functions can escribe some complicate curves! y 2 1 x 2 + 2xy y 3 = 0 3 2 1 1 1 2 x Each place on the curve shown here has a well efine slope (graient) Therefore, we shoul be able to use calculus to figure out the graient function for the implicit function. x Implicit Differentiation 3/ 11 Arian Jannetta
Applying the chain rule to implicit functions Consier the following two erivatives x (x2 ) an x (y2 ) As we have seen, the first is just 2x. Can the secon be just 2y? That woul be the answer if we were ifferentiating with respect to y, but the x tells us that that we are still ifferentiating with respect to x. Since y 2 epens on x we have to use the chain rule. Let u=y 2 so that x (y2 ) = u x Applying the chain rule we can write: an u = 2y x (y2 ) = u x = u = 2y x x This time the answer inclues the graient term x! Implicit Differentiation 4/ 11 Arian Jannetta
Implicit ifferentiation Implicit ifferentiation Given F(y) where y is a function of x, then we can ifferentiate with respect to x using: F(y) = F(y) x So, ifferentiate with respect to y an then multiply by x. Here are some examples: x x (y3 )= (y3 ) = 3y2 x x x (siny)= (cos y) = cos y x x x (5ey )= (5ey ) = 5ey x x Implicit Differentiation 5/ 11 Arian Jannetta
Graient of a circle A circle is efine by the equation x 2 + y 2 = 25 Fin the graient of the circle at the point (3,4). We ifferentiate each term with respect to x x (x2 )+ x (y2 )= x (25) Remember the rule for ifferentiating y; it follows the usual rule but you must also multiply by x : 2x+2y x = 0 Rearrange to get the graient: x = 2x 2y = x y At the point (3,4) the graient of the circle is 3 4. Implicit Differentiation 6/ 11 Arian Jannetta
Differentiating an implicit function A curve is efine by the equation Fin an expression for x. Differentiating term by term x 3 + 2y 2 3x+4y 19=0 3x 2 + 4y x 3+4 x = 0 Factorise the terms with x : x (4y+ 4)+3x2 3=0 Now rearrange to isolate the x term: x = 3 3x2 4y+ 4 Implicit Differentiation 7/ 11 Arian Jannetta
The rules of ifferentiation The rules of ifferentiation - the chain, prouct an quotient rules - work in the usual way. The only ifference occurs when ifferentiating terms containing y, in which case we have to o it as shown in the previous examples. Incorporating the chain rule Fin the graient function of the curve which is escribe by ln(y 2 + 3) 2x 2 = 5 As before we ifferentiate term by term. The first must be foun using the chain rule. 2y y 2 + 3 x 4x=0 Now rearrange to get the graient: x = 4x(y2 + 3) = 2x(y2 + 3) 2y y Implicit Differentiation 8/ 11 Arian Jannetta
Another example; the prouct rule Consier the curve efine by Fin for this curve. x sin y+ 3x 2 y 3 + 1=0 Differentiate term by term. In this case we will nee to use the prouct rule with the secon term. cosy x + 3x2 (3y 2 x )+y3 (6x)= 0 Now simplify the terms cosy x + 9x2 y 2 x + 6xy3 = 0 Factorise: x (cosy+ 9x2 y 2 )+6xy 3 = 0 an rearrange to get the graient: x = 6xy 3 cosy+ 9x 2 y 2 Implicit Differentiation 9/ 11 Arian Jannetta
Cutting own the working out... In this type of work it is useful to use the prime notation. You might fin it easier (an quicker) to o these questions by writing y instea of x. Incorporating the quotient rule Fin an expression for the graient for the curve efine by 2tan y+ 3x 5y 2= 1 Differentiate term by term. The quotient rule is applie to the secon term on the LHS. (2sec 2 y)y + (5y2 )(3) 3x(10yy ) (5y 2 ) 2 = 0 Simplify: (2sec 2 y)y + 15y2 30xyy 25y 4 = 0 (2sec 2 y)y 3y 6xy + = 0 5y 3 Cross multiply by 5y 3 : (10y 3 sec 2 y)y + 3y 6xy = 0 Now factorise (10y 3 sec 2 y 6x)y + 3y= 0 Rearrange to get the graient: y 3y = 10y 3 sec 2 y 6x Implicit Differentiation 10/ 11 Arian Jannetta
Test yourself... You shoul be able to solve the following problems if you have unerstoo everything in these notes. 1 Fin x y 2 + e 2y 2 Given the curve x 2 y 2 + 7x=2, fin an expression for x. 3 Given the curve 2x 3 y 2 siny= 0, fin an expression for x. 4 Fin the equation of the normal to the curve x 2 4xy+ y 2 = 24 at the point (2,10). Answers: 1 2 x (y+ e2y ) 2 x = 2x+7 2y 3 x = 6x 2 y 2 4x 3 y cosy 4 Normal equation is y= 1 32 3 x+ 3 Implicit Differentiation 11/ 11 Arian Jannetta