5Example 2 Find derivatives for A) f (x) = ex / 2 f ’(x) = ex / 2 B) f (x) = 2ex +x2 f ’(x) = 2ex + 2xC) f (x) = -7xe – 2ex + e2 f ’(x) = -7exe-1 – 2exRemember that e is a real number, so the power rule is used to find the derivative of xe.Also e2 is a constant, so its derivative is 0.
6Review is equivalent to Domain: (0, ∞)Range: (-∞, ∞)Range: (0, ∞)Domain: (-∞, ∞)* These are inverse function. The graphs are symmetric with respect to the line y=x* There are many different bases for a logarithmic functions. Two special logarithmic functions are common logarithm (log10x or log x) and natural logarithm (logex = ln x)
8Use the limit definition to find the derivative of ln x Optional slide:Use the limit definition to find the derivative of ln xFindProperty 2Multiply by 1 which is x / xSet s = h / xSo when h approaches 0, s also approaches oProperty 3Definition of eProperty 4: ln(e)=1
9The Derivative of ln xTherefore: The derivative of f (x) = ln x is f ’(x) =
13Example 5 An Internet store sells blankets. If the price-demand equation is p = 200(0.998)x, find the rate of change of pricewith respect to demand when the demand is 400 blanketsand explain the result.p’ = 200 (.998)x ln(0.998)p’(400) = 200 (.998)400 ln(0.998) =When the demand is 400 blankets, the price is decreasing about 18 cents per blanket
14Example 6 t = 30 corresponds to 2010 C(30) = 83 – 9 ln30 = 52.4 A model for newspaper circulation is C(t) = 83 – 9 ln twhere C is newspaper circulation (in millions) and t is the number ofyears (t=0 corresponds to 1980). Estimate the circulation and find therate of change of circulation in 2010 and explain the result.t = 30 corresponds to 2010C(30) = 83 – 9 ln30 = 52.4C(t)’ = C’(30) =The circulation in 2010 is about 52.4 million and is decreasing at the rate of 0.3 million per year
15Example 7: Find the equation of the tangent line to the graph of f = 2ex + 6x at x = 0 Y = mx + bf’(x) = 2ex + 6m = f’(0) = 2(1) + 6 = 8y=f (0) = 2(1) + 6(0) = 22 = 8(0) + b so b = 2The equation is y = 8x + 2
16Example 8: Use graphing calculator to find the points of intersection F(x) = (lnx)2 and g(x) = xOn your calculator, press Y=Type in the 2 functions above for Y1 and Y2Press ZOOM, 6:ZStandardTo have a better picture, go back to ZOOM, 2: Zoom In*Now, to find the point of intersection (there is only 1 in this problem), press 2ND, TRACE then 5: intersectPlay with the left and right arrow to find the linking dot, when you see it, press ENTER, ENTER again, then move it to the intersection, press ENTER. From there, you should see the point of intersection( , )