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# Lesson 16: Implicit Differentiation

## by Matthew Leingang, Clinical Associate Professor of Mathematics at New York University on Nov 02, 2007

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## Lesson 16: Implicit DifferentiationPresentation Transcript

• Lesson 16 (Section 3.6) Implicit Diﬀerentiation Derivatives of Inverse Functions Math 1a October 31, 2007 Announcements OH: Mondays 1–2, Tuesdays 3–4, Wednesdays 1–3 (SC 323)
• Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5).
• Problem Find the slope of the line which is tangent to the curve x2 + y2 = 1 at the point (3/5, −4/5). Solution (Old Way) y2 = 1 − x2 =⇒ y = − 1 − x 2 −2x dy x =− √ =√ dx 2 1 − x2 2 1−x dy 3/5 3/5 3 = = =. dx x=3/5 4/5 4 1 − (3/5)2
• Solution (New Way) Diﬀerentiating this, but pretend that y is a function of x. We have: dy 2x + 2y = 0, dx or dy x =− . dx y 3 4 So if x = , y = − , and r = 1, then the slope of the line tangent 5 5 to the circle is dy 3/5 3 = =. dx ( 3 ,− 4 ) 4/5 4 5 5
• Summary any time a relation is given between x and y , we may diﬀerentiate y as a function of x even though it is not explicitly deﬁned This is called implicit diﬀerentiation.
• Example Find the equation of the line tangent to the curve y 2 = x 2 (x + 1) = x 3 + x 2 at the point (3, −6).
• Example Find the equation of the line tangent to the curve y 2 = x 2 (x + 1) = x 3 + x 2 at the point (3, −6). Solution Diﬀerentiating the expression implicitly with respect to x dy = 3x 2 + 2x, so gives 2y dx 3x 2 + 2x dy = , and dx 2y 3 · 32 + 2 · 3 dy 11 =− . = dx 2(−6) 4 (3,−6)
• Example Find the equation of the line tangent to the curve y 2 = x 2 (x + 1) = x 3 + x 2 at the point (3, −6). Solution Diﬀerentiating the expression implicitly with respect to x dy = 3x 2 + 2x, so gives 2y dx 3x 2 + 2x dy = , and dx 2y 3 · 32 + 2 · 3 dy 11 =− . = Thus the equation of the dx 2(−6) 4 (3,−6) tangent line is 11 y +6=− (x − 3). 4
• Example Find the horizontal tangent lines to the same curve: y 2 = x 3 + x 2
• Example Find the horizontal tangent lines to the same curve: y 2 = x 3 + x 2 Solution 1 We need 3x 2 + 1 = 0, or x = √ . Then 3 1 1 y2 = √ + , 33 3 Thus 1 1 +√ y =± 3 33
• An example from chemistry Example The van der Waals equation describes nonideal properties of a gas: n2 (V −nb) = nRT , P +a V2 where P is the pressure, V the volume, T the temperature, n the number of moles of the gas, R a constant, a is a measure of attraction between particles of the gas, and b a measure of particle size.
• An example from chemistry Example The van der Waals equation describes nonideal properties of a gas: n2 (V −nb) = nRT , P +a V2 where P is the pressure, V the volume, T the temperature, n the number of moles of the gas, R a constant, a is a measure of attraction between particles of the gas, and b a measure of particle size.
• Deﬁnition The isothermic compressibility of a ﬂuid is deﬁned by dV 1 β=− dP V with temperature held constant.
• Deﬁnition The isothermic compressibility of a ﬂuid is deﬁned by dV 1 β=− dP V with temperature held constant. The smaller the β, the “harder” the ﬂuid.
• Let’s ﬁnd the compressibility of a van der Waals gas. Diﬀerentiating the van der Waals equation by treating V as a function of P gives an2 2an2 dV dV + (V − bn) 1 − P+ = 0, V2 V 3 dP dP
• Let’s ﬁnd the compressibility of a van der Waals gas. Diﬀerentiating the van der Waals equation by treating V as a function of P gives an2 2an2 dV dV + (V − bn) 1 − P+ = 0, V2 V 3 dP dP so V 2 (V − nb) 1 dV β=− = 2abn3 − an2 V + PV 3 V dP
• Let’s ﬁnd the compressibility of a van der Waals gas. Diﬀerentiating the van der Waals equation by treating V as a function of P gives an2 2an2 dV dV + (V − bn) 1 − P+ = 0, V2 V 3 dP dP so V 2 (V − nb) 1 dV β=− = 2abn3 − an2 V + PV 3 V dP What if a = b = 0?
• Let’s ﬁnd the compressibility of a van der Waals gas. Diﬀerentiating the van der Waals equation by treating V as a function of P gives an2 2an2 dV dV + (V − bn) 1 − P+ = 0, V2 V 3 dP dP so V 2 (V − nb) 1 dV β=− = 2abn3 − an2 V + PV 3 V dP What if a = b = 0? dβ What is the sign of ? db
• Let’s ﬁnd the compressibility of a van der Waals gas. Diﬀerentiating the van der Waals equation by treating V as a function of P gives an2 2an2 dV dV + (V − bn) 1 − P+ = 0, V2 V 3 dP dP so V 2 (V − nb) 1 dV β=− = 2abn3 − an2 V + PV 3 V dP What if a = b = 0? dβ What is the sign of ? db dβ What is the sign of ? da
• Theorem (The Inverse Function Theorem) Let f be diﬀerentiable at a, and f (a) = 0. Then f −1 is deﬁned in a neighborhood of b = f (a), and 1 (f −1 ) (b) = (f −1 (b)) f
• Theorem (The Inverse Function Theorem) Let f be diﬀerentiable at a, and f (a) = 0. Then f −1 is deﬁned in a neighborhood of b = f (a), and 1 (f −1 ) (b) = (f −1 (b)) f “Proof”. If y = f (x), take f −1 of both sides to get x = f −1 (f (x)), so 1 = (f −1 ) (f (x))f (x).
• Using implicit diﬀerentiation to ﬁnd derivatives Example √ dy Find if y = x. dx
• Using implicit diﬀerentiation to ﬁnd derivatives Example √ dy Find if y = x. dx Solution √ If y = x, then y 2 = x, so dy dy 1 1 = √. 2y = 1 =⇒ = dx dx 2y 2x
• The power rule for rational numbers Example dy if y = x p/q , where p and q are integers. Find dx
• The power rule for rational numbers Example dy if y = x p/q , where p and q are integers. Find dx Solution We have yq = xp dy qy q−1 = px p−1 dx p x p−1 dy = · q−1 dx qy
• The power rule for rational numbers Example dy if y = x p/q , where p and q are integers. Find dx Solution We have yq = xp dy qy q−1 = px p−1 dx p x p−1 dy = · q−1 dx qy Now y q−1 = x p(q−1)/q = x p−p/q so x p−1 = x p−1−(p−p/q) = x p/q−1 y q−1
• Arcsin Remember that 1 arcsin x = sin−1 x = sin x
• Arcsin Remember that 1 arcsin x = sin−1 x = sin x Example Find the derivative of arcsin.
• Arcsin Remember that 1 arcsin x = sin−1 x = sin x Example Find the derivative of arcsin. Solution We have y = arcsin x =⇒ x = sin y , so dy dy 1 1 = cos y =⇒ = . dx dx cos y We would like to express this in terms of x. Since x = sin y , we have d 1 cos y = 1 − x 2 =⇒ arcsin x = √ . dx 1 − x2
• Arccos Example Find the derivative of arccos.
• Arccos Example Find the derivative of arccos. Answer −1 d (arccos x) = √ dx 1 − x2
• Arctan Example Find the derivative of arctan.
• Arctan Example Find the derivative of arctan. Solution From y = arctan x we have dy x = tan y =⇒ 1 = sec2 y , dx so dy 1 = . sec2 y dx
• Arctan Example Find the derivative of arctan. Solution From y = arctan x we have dy x = tan y =⇒ 1 = sec2 y , dx so dy 1 = . sec2 y dx But for all y dy 1 1 + tan2 y = sec2 y , =⇒ = . 1 + x2 dx
• Arcsec Example Find the derivative of arcsec.
• Arcsec Example Find the derivative of arcsec. Solution If y = arcsec x then x = sec y , so dy 1 = sec y tan y . dx Thus dy 1 1 =√ = . dx x tan y x x2 − 1