MIT OpenCourseWare http://ocw.mit.edu 18.01 Single Variable Calculus Fall 2006 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � Lecture 7 18.01 Fall 2006 Lecture 7: Continuation and Exam Review Hyperbolic Sine and Cosine Hyperbolic sine (pronounced “sinsh”): sinh(x) = ex − e−x 2 Hyperbolic cosine (pronounced “cosh”): ex + e−x cosh(x) = 2 x xd sinh(x) = d e − e−x = e − (−e−x) = cosh(x)dx dx 2 2 Likewise, d cosh(x) = sinh(x)dx d(Note that this is different from cos(x).)dx Important identity: cosh2(x) − sinh2(x) = 1 Proof: � �2 � x �2 cosh2(x) − sinh2(x) = ex +2 e−x − e −2 e−x 1 � � 1 � � 1cosh2(x) − sinh2(x) = 4 e 2x + 2e x e−x + e−2x − 4 e 2x − 2 + e−2x = 4(2 + 2) = 1 Why are these functions called “hyperbolic”? Let u = cosh(x) and v = sinh(x), then u 2 − v 2 = 1 which is the equation of a hyperbola. Regular trig functions are “circular” functions. If u = cos(x) and v = sin(x), then u 2 + v 2 = 1 which is the equation of a circle. 1� �� �Lecture 7 18.01 Fall 2006 Exam 1 Review General Differentiation Formulas (u + v)� = u� + v� (cu)� = cu� (uv)� = u�v + uv� (product rule) u � = u�v − uv� (quotient rule) v v2 d f(u(x)) = f�(u(x)) u�(x) (chain rule) dx · You can remember the quotient rule by rewriting u � = (uv−1)� v and applying the product rule and chain rule. Implicit differentiation Let’s say you want to find y� from an equation like y 3 + 3xy 2 = 8 dInstead of solving for y and then taking its derivative, just take of the whole thing. In this dx example, 3y 2 y� + 6xyy� + 3y 2 = 0 (3y 2 + 6xy)y� = −3y 2 y� = −3y2 3y2 + 6xy Note that this formula for y� involves both x and y. Implicit differentiation can be very useful for taking the derivatives of inverse functions. For instance, y = sin−1 x sin y = x ⇒ Implicit differentiation yields (cos y)y� = 1 and 1 1 y� = = cos y √1 − x2 2� � Lecture 7 18.01 Fall 2006 Specific differentiation formulas You will be responsible for knowing formulas for the derivatives and how to deduce these formulas n xfrom previous information: x , sin−1 x, tan−1 x, sin x, cos x, tan x, sec x, e , ln x . dFor example, let’s calculate sec x: dx d d 1 −(− sin x)sec x = = = tan x sec x dx dx cos x cos2 x d dYou may be asked to find sin x or cos x, using the following information: dx dx sin(h)lim = 1 h 0 h→lim cos(h) − 1 = 0 h 0 h→Remember the definition of the derivative: df(x) = lim f(x + Δx) − f(x) dx Δx 0 Δx→Tying up a loose end dHow to find x r, where r is a real (but not necessarily rational) number? All we have done so far dx is the case of rational numbers, using implicit differentiation. We can do this two ways: 1st method: base e x = e ln x x r = � e ln x �r = e r ln x d dx x r = d dx e r ln x = e r ln x d dx (r ln x) = e r ln x r x d dx x r = x r � r x � = rx r−1 2nd method: logarithmic differentiation f�(ln f)� = f f = x r ln f = r ln x r(ln f)� = x f� = f(ln f )� = x r r = rx r−1 x 3� � �� Lecture 7 18.01 Fall 2006 Finally, in the first lecture I promised you that you’d learn to differentiate anything— even something as complicated as dx tan−1 x e dx So let’s do it! d d e uv = e uv (uv) = e uv (u�v + uv�)dx dx Substituting, de x tan−1 x = e x tan−1 x tan−1 x + x 1 dx 1 + x2
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