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Implicit DifferentiationMath165: Business CalculusRoy M. LowmanSpring 2010Roy M. Lowman Implicit DifferentiationImplicit Differentiation: prerequisitesTo understand implicit differentiation it is necessary tounderstand the chain rule.Given: y = y(x), y is a function of xSimple Power Rule:ddxxn= nxn−1General Power Rule:ddxy(x)n= nyn−1· y0(x) , due to chainrule:ddxyn=ddyyn·dydx= nyn−1· y0(x)ddyyn= nyn−1is not the same asddxyn= nyn−1· y0(x)Inddyyn= nyn−1the variable of differentiation is y (i.e.ddy)the same as the variable in ynso the simple power rule is used.Inddxyn= nyn−1· y0(x) the variable of differentiation is x(i.e.ddx) not the same as the variable in ynso the generalpower rule is used.Roy M. Lowman Implicit DifferentiationImplicit Differentiation: prerequisitesTo understand implicit differentiation it is necessary tounderstand the chain rule.Given: y = y(x), y is a function of xSimple Power Rule:ddxxn= nxn−1General Power Rule:ddxy(x)n= nyn−1· y0(x) , due to chainrule:ddxyn=ddyyn·dydx= nyn−1· y0(x)ddyyn= nyn−1is not the same asddxyn= nyn−1· y0(x)Inddyyn= nyn−1the variable of differentiation is y (i.e.ddy)the same as the variable in ynso the simple power rule is used.Inddxyn= nyn−1· y0(x) the variable of differentiation is x(i.e.ddx) not the same as the variable in ynso the generalpower rule is used.Roy M. Lowman Implicit DifferentiationImplicit Differentiation: prerequisitesTo understand implicit differentiation it is necessary tounderstand the chain rule.Given: y = y(x), y is a function of xSimple Power Rule:ddxxn= nxn−1General Power Rule:ddxy(x)n= nyn−1· y0(x) , due to chainrule:ddxyn=ddyyn·dydx= nyn−1· y0(x)ddyyn= nyn−1is not the same asddxyn= nyn−1· y0(x)Inddyyn= nyn−1the variable of differentiation is y (i.e.ddy)the same as the variable in ynso the simple power rule is used.Inddxyn= nyn−1· y0(x) the variable of differentiation is x(i.e.ddx) not the same as the variable in ynso the generalpower rule is used.Roy M. Lowman Implicit DifferentiationImplicit Differentiation: prerequisitesTo understand implicit differentiation it is necessary tounderstand the chain rule.Given: y = y(x), y is a function of xSimple Power Rule:ddxxn= nxn−1General Power Rule:ddxy(x)n= nyn−1· y0(x) , due to chainrule:ddxyn=ddyyn·dydx= nyn−1· y0(x)ddyyn= nyn−1is not the same asddxyn= nyn−1· y0(x)Inddyyn= nyn−1the variable of differentiation is y (i.e.ddy)the same as the variable in ynso the simple power rule is used.Inddxyn= nyn−1· y0(x) the variable of differentiation is x(i.e.ddx) not the same as the variable in ynso the generalpower rule is used.Roy M. Lowman Implicit DifferentiationImplicit Differentiation: prerequisitesTo understand implicit differentiation it is necessary tounderstand the chain rule.Given: y = y(x), y is a function of xSimple Power Rule:ddxxn= nxn−1General Power Rule:ddxy(x)n= nyn−1· y0(x) , due to chainrule:ddxyn=ddyyn·dydx= nyn−1· y0(x)ddyyn= nyn−1is not the same asddxyn= nyn−1· y0(x)Inddyyn= nyn−1the variable of differentiation is y (i.e.ddy)the same as the variable in ynso the simple power rule is used.Inddxyn= nyn−1· y0(x) the variable of differentiation is x(i.e.ddx) not the same as the variable in ynso the generalpower rule is used.Roy M. Lowman Implicit DifferentiationImplicit Differentiation: prerequisitesTo understand implicit differentiation it is necessary tounderstand the chain rule.Given: y = y(x), y is a function of xSimple Power Rule:ddxxn= nxn−1General Power Rule:ddxy(x)n= nyn−1· y0(x) , due to chainrule:ddxyn=ddyyn·dydx= nyn−1· y0(x)ddyyn= nyn−1is not the same asddxyn= nyn−1· y0(x)Inddyyn= nyn−1the variable of differentiation is y (i.e.ddy)the same as the variable in ynso the simple power rule is used.Inddxyn= nyn−1· y0(x) the variable of differentiation is x(i.e.ddx) not the same as the variable in ynso the generalpower rule is used.Roy M. Lowman Implicit DifferentiationImplicit Differentiation: prerequisitesTo understand implicit differentiation it is necessary tounderstand the chain rule.Given: y = y(x), y is a function of xSimple Power Rule:ddxxn= nxn−1General Power Rule:ddxy(x)n= nyn−1· y0(x) , due to chainrule:ddxyn=ddyyn·dydx= nyn−1· y0(x)ddyyn= nyn−1is not the same asddxyn= nyn−1· y0(x)Inddyyn= nyn−1the variable of differentiation is y (i.e.ddy)the same as the variable in ynso the simple power rule is used.Inddxyn= nyn−1· y0(x) the variable of differentiation is x(i.e.ddx) not the same as the variable in ynso the generalpower rule is used.Roy M. Lowman Implicit DifferentiationImplicit Differentiation: prerequisitesTo understand implicit differentiation it is necessary tounderstand the chain rule.Given: y = y(x), y is a function of xSimple Power Rule:ddxxn= nxn−1General Power Rule:ddxy(x)n= nyn−1· y0(x) , due to chainrule:ddxyn=ddyyn·dydx= nyn−1· y0(x)ddyyn= nyn−1is not the same asddxyn= nyn−1· y0(x)Inddyyn= nyn−1the variable of differentiation is y (i.e.ddy)the same as the variable in ynso the simple power rule is used.Inddxyn= nyn−1· y0(x) the variable of differentiation is x(i.e.ddx) not the same as the variable in ynso the generalpower rule is used.Roy M. Lowman Implicit Differentiationexplicit vs implicit functiony3= 3x4+ x, gives y as a function of x in implicit form.Given a set of x values, you can solve for the corresponding yvalues , plot the points (x, y) and construct a graph of thefunction y = y(x).The relation gives the function y = y(x) but in implicit form.In this case, it is possible to explicitly solve for y(x). This willgive the same function y = y(x) but in explicit form (the onewe usually use).y3= 3x4+ x (1)(y3)13= (3x4+ x)13(2)y(x) = (3x4+ x)13(3)Roy M. Lowman Implicit Differentiationexplicit vs implicit functiony3= 3x4+ x, gives y as a function of x in implicit form.Given a set of x values, you can solve for the corresponding yvalues , plot the points (x, y) and construct a graph of thefunction y = y(x).The relation gives the function y = y(x) but in implicit form.In this case, it is possible to explicitly solve for y(x). This willgive the same function y = y(x) but in explicit form (the onewe usually use).y3= 3x4+ x (1)(y3)13= (3x4+ x)13(2)y(x) = (3x4+ x)13(3)Roy M. Lowman Implicit Differentiationexplicit vs implicit functiony3= 3x4+ x, gives y as a function of x in implicit form.Given a set of x values, you can solve for the corresponding yvalues , plot the points (x, y) and

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