Understanding Implicit Differentiation: Examples and Techniques
Explore the concept of implicit differentiation, including its application to single terms and functions with mixed x and y terms. Learn key examples and techniques to find gradient functions effectively.
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Differentiation: Implicit KUS objectives BAT Understand Implicit differentiation and find the gradient function for functions in x and y Starter differentiate: ?3?+ ?2 2 ?3+ 32 6?2(?3+ 3) 2 ?3?+ ?2 (3?3? + 2?) ?sin? ?2?2?2 54 2?2?2?2 54+ 8? ?2?2?2 53 sin? + ? cos? ?3 4?(1 + 3?2) tan? 3 ?2tan? ?3sec2? tan2? !!!!
Notes Implicit differentiation of single terms For some equations it is impossible to rearrange to give y = f(x) and hence differentiate. One approach is to use the chain rule to differentiate each term without rearrangement For example differentiate y2 ? ?2 ?? ?? ? ??(?2) = ??= 2? ?? ?? ?? ?? ? Differentiate the y term as you would for an x term And multiply by ?? More examples: 8?? ?? ?? 8? 6??? 3?2 ??
WB30a One approach is to use the chain rule to differentiate each term without rearrangement. Here are some Key examples: ? ?? ??= ??? 1?? ?? Term Gradient function Term Gradient function ?4 2? + 5?4.?? ?2+ ?5 4?3 .?? ?? ?? 1 ? 1 ?3 ?2 .?? ?? ?? cos? .?? 1 3? 4 .?? ?? ?? 1 2 3? 1 ?????2?.?? 3 .?? ? 3?2 2 ? ?? sin? cot? ?? ?? 1 6? 3 1 2? 4 .?? ?? .?? ?? 1 ? .?? ?? 1 7(? + 1) 8 .?? ? + 17 ?? ?? ?? 6? 6? .?? ? 6? ?? ?2 3 .?? 2? ln ?2 3 ln? ??
Notes Implicit differentiation of functions Now we can differentiate an entire function with mixed terms in x and y And attempt to rearrange the result to find a gradient function For example the equation of this circle 2? + 2??? ?2+ ?2= 3? + 7 ??= 3 By differentiating each term implicitly Then rearranging ?? ??=3 2? We get a gradient function in terms of x and y 2?
? ?? ??= ??? 1?? WB31 Differentiate the given functions ?? function Gradient function ?2= 6?4 ??=12?3 2??? ??= 24?3 or ?? ? ?2+ ?2+ 6? = 25 ?? ??+ 6 = 0 or ?? ??= (?+3) 2? + 2? ? ??? ? = 4? + 7 ?? ??= 4 or ?? cos? ??= 4sec? ?2= sec? ?? ??= sec? tan? ?? ??= 1 2? ?? 2?sec?tan? ??= ?? 3 ???? ?? ??= ?? ? ??= ?? 2?2??? or ?2?=1 3?3+1 2?2+ 8 ??= ?2+ ? 1 ? + 2 2? ??? = ln(? + 2) ?2 1 ? ?? ??=
The curve C has equation ?3+ ? + ?3+ 3? = 6 a) Find dy/dx in terms of x and y b) Hence find the gradient of C at the point (1, 1) WB32 ?) 3?2+ 1 + 3?2?? ??+ 3?? ??= 0 ?? ?? 3?2+ 3 = 3?2 1 You now have a formula for the gradient, but in terms of x AND y, not just x ?? ??= 3?2 1 3?2+ 3 b) at the point (1, 1) 3(1)2 1 3(1)2+3 ?? ??= = 2 3
Notes Implicit differentiation of functions A special case is the derivative of a product of x and y Key example: differentiate ?(?,?) = ?? using the Product rule ? .?? ? ??(??) = ??? ??+ ? ??+ ?.1 = Key example: differentiate ?(?,?) = ??2using the Product rule ? .2??? ? ??(??^2) = 2???? ??+ ?2 ??+ ?^2.1 =
WB33abc Differentiate the given functions (including products) function Gradient function 2?2+ ?? ? = 2 4? + ? + ??? ?? ??= 0 ?? ?? ??= 4? ? ? 1 ?2+ 2?? + ?2= 7 2? + 2? + 2??? + 2??? ??= 0 ?? ?? ??= 2? 2? 2? + 2? = 1 2? ?3+ 3?2? = 0 2?3+ 2? .3?2?? + 6?? + 3?2 .?? = 0 ?? ?? ?? ??= 2?3 6?? 6?? + 3?2
WB33de Differentiate the given functions (including products) function Gradient function 1.?? ? + ? .1 ?? ?? = 0 ?ln? = 10 ? ?? ??= ?ln? ? ?? ??= 2? .??+ ?2 .???? ? = ?2?? ?? 2? ?? 1 ?2 ?? ?? ??=
WB33fg Differentiate the given functions (including products) function Gradient function ?? ?? 1.sin2? + ?.2cos2? = 0 ?sin2? = 1 ?? ??= sin2? 2? cos2?= 1 2?tan2? 2? .tan? + ?2 .sec2??? ?2tan? = cos? = sin? ?? ?? ??= sin? 2?tan? ?2sec2?
WB34The curve C has equation 4??2 5? = 11 a) Find dy/dx in terms of x and y b) Hence find the gradient of C at the point (1, 2) 4 .?2+ 4? .2??? ?) 5 = 0 ?? ?? ?? 8?? = 5 4 ?2 ?? ??=5 4 ?2 You now have a formula for the gradient, but in terms of x AND y, not just x 8?? b) at the point (1, 2) ?? ??= 11 16
WB35 A circle has equation ?2+ ?2= 25 Find two equations of tangents to the curve when x = 4 differentiating2? + 2??? ?? = 0 rearranges to ?? ?? = ? ? ?2= 25 42 when ? = 4 ? = 3 ?? ?? = 4 At (4, 3) 3 ?? ?? =4 At (4, -3) 3 y 3 = 4 3(? 4) y + 3 =4 3(? 4) Tangent 4? + 3? = 25 Tangent 4? 3? = 25
WB36 Exam Q A curve C is described by the equation 3?2 4?2+ 4? 5? + 12 = 0 Find an equation of the normal to C at point (3, 3), giving your answer in the form ?? + ?? + ? = 0 6? 8??? ??+ 4 5?? M1A1 ??= 0 at the point (3, 3) ?? ?? =6?+4 8?+5=22 M1A1 29 Gradient of normal is ? = 29 A1 22 Equation of normal is ? 3 = 29 22? 3 29? + 22? 153 = 0 M1A1
WB37 Find the value of dy/dx at the point (1,1) where ?2?ln? = ? + ? 2 Give your answer in terms of e 2?2? .ln? + ?2? .1 ?? ?? = 1 +?? ? ?? at the point (1, 1) 2?2 .ln1 + ?2 .1 ?? ?? = 1 +?? 1 ?? Rearranges to dy dx?2 1 = 1 2?2ln1 But ln 1 = 0 so dy dx?2 1 = 1 dy dx = 1 ?2 1
KUS objectives BAT Understand Implicit differentiation and find gradient function for functions in x and y Crucial points Make sure that you understand the process of differentiating an equation implicitly Write one thing you have learned Write one thing you need to improve
