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# Lesson 16: Implicit Differentiation

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Lesson 16: Implicit Differentiation
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Lesson 16 (Section 3.6)Implicit DiﬀerentiationDerivatives of Inverse FunctionsMath 1aOctober 31, 2007AnnouncementsOH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)Solution (Old Way)y 2 = 1 − x 2=⇒ y = −1 − x 2dy−2xx= − √= √dx2 1 − x 21 − x 2dy3/53/53===.dx4/54x =3/51 − (3/5)2ProblemFind the slope of the line which is tangent to the curvex 2 + y 2 = 1at the point (3/5, −4/5).ProblemFind the slope of the line which is tangent to the curvex 2 + y 2 = 1at the point (3/5, −4/5).Solution (Old Way)y 2 = 1 − x 2=⇒ y = −1 − x 2dy−2xx= − √= √dx2 1 − x 21 − x 2dy3/53/53===.dx4/54x =3/51 − (3/5)2Solution (New Way)Diﬀerentiating this, but pretend that y is a function of x . We have:dy2x + 2y= 0,dxordyx= − .dxy34So if x =, y = − , and r = 1, then the slope of the line tangent55to the circle isdy3/53==.dx (34/54,− 4 )55Summaryany time a relation is given between x and y , we may diﬀerentiatey as a function of x even though it is not explicitly deﬁnedThis is called implicit diﬀerentiation.SolutionDiﬀerentiating the expressionimplicitly with respect to xdygives 2y= 3x 2 + 2x , sodxdy3x 2 + 2x=, anddx2ydy3 · 32 + 2 · 311== −.dx2(−6)4Thus the equation of the(3,−6)tangent line is11y + 6 = −(x − 3).4ExampleFind the equation of the line tangent to the curvey 2 = x 2(x + 1) = x 3 + x 2at the point (3, −6).Thus the equation of thetangent line is11y + 6 = −(x − 3).4ExampleFind the equation of the line tangent to the curvey 2 = x 2(x + 1) = x 3 + x 2at the point (3, −6).SolutionDiﬀerentiating the expressionimplicitly with respect to xdygives 2y= 3x 2 + 2x , sodxdy3x 2 + 2x=, anddx2ydy3 · 32 + 2 · 311== −.dx2(−6)4(3,−6)ExampleFind the equation of the line tangent to the curvey 2 = x 2(x + 1) = x 3 + x 2at the point (3, −6).SolutionDiﬀerentiating the expressionimplicitly with respect to xdygives 2y= 3x 2 + 2x , sodxdy3x 2 + 2x=, anddx2ydy3 · 32 + 2 · 311== −.dx2(−6)4Thus the equation of the(3,−6)tangent line is11y + 6 = −(x − 3).4Solution1We need 3x 2 + 1 = 0, or x = √ . Then311y 2 = √ +,3 33Thus11y = ±+√333ExampleFind the horizontal tangent lines to the same curve: y 2 = x 3 + x 2ExampleFind the horizontal tangent lines to the same curve: y 2 = x 3 + x 2Solution1We need 3x 2 + 1 = 0, or x = √ . Then311y 2 = √ +,3 33Thus11y = ±+√333Document Outline
• Announcements
• The big idea, by example
• Examples
• Inverse Functions
• The power rule for rational powers
• Derivatives of inverse trigonometric functions

Lesson 16: Implicit Differentiation

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