102 Problems Calculus AB Students Should Know: Solutions. 18. product rule d. 19. d sin x. 20. chain rule d e 3x2) = e 3x2 ( 6x) = 6xe 3x2

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1 Problems Calculus AB Stuents Shoul Know: Solutions. + ) = + =. chain rule ) e = e = e. ) =. ) = ln ) = + = = +. ln ) =. ) log ) =. sin ) = cos. cos ) = sin. tan ) = sec. cot ) = csc. sec ) = sec tan. csc ) = csc cot.. sin ) = + + ) + = + ) = = ) ln = ln ln tan ) = +. prouct rule ln ) = ln + ) = ln +. prouct rule e cos ) = e cos + e sin ) = e cos e sin. ) sin + cos ) cos sin sin ) = = + cos + cos ) cos + cos + sin = cos + + cos ) + cos ) = quotient rule + cos. chain rule e ) = e ) = e. chain rule ) + ) = + ) = + ). chain rule ) = ) ) =. funamental theorem of calculus sin t t = sin. chain rule sin t t = sin ) = sin. sin t t = sin. f ) = slope at = = =. f ) =, instantaneous rate of change = f ) =. f) f) = = Copyright c by The Regents of the University of California

2 Problems Calculus AB Stuents Shoul Know: Solutions. a) f ) = = ) = )+) = so local maimum at = with maimum value f ) = by first erivative test increasing then ecreasing) or secon erivative test f ) = < concave own) an local minimum at = with minimum value f) = by first erivative test ecreasing then increasing) or secon erivative test f ) = > concave up) b) f ) > for < increasing f ) < for < < ecreasing f ) > for > increasing c) f ) = = ) f ) > for > concave up f ) < for < concave own ) inflection point at = coorinates, ) where concavity changes from own to up or vice versa. f) =, f ) =, slope f ) =, perpenicular slope, tangent line y = ) + =, normal line y = ) + = +. implicit ifferentiation + yy = plug in =, y = to get + y = so y = tangent line y = ) + = +. f) =, f ) = e, f ) =, tangent line y =, plug in = to get f ) correct to, while e gives wrong answer on a calculator). implicit ifferentiation of A = πr A = πrr = π)) = π cm /s. implicit ifferentiation C = A B +AB = ))+)) =. velocity vt) = t) = t + acceleration at) = v t) =. Let an y be the length an with of the enclose area, then fencing neee is + y = so y =. The area A = y = ) =. Maimize: A = =, =, y = =. The rectangle that enclose the largest area is a square with length m an with of m.. The area of a circle of raius r is A = πr. The circumference of a circle of raius r is C = πr. The area of a triangle with base b an height h is A = bh. The area of a trapezoi with bases a an b an height h is A = a + b)h. The volume of a sphere of raius r is V = πr. The surface area of a sphere of raius r is SA = πr. The volume of a cyliner/prism with base area A an height h is V = Ah. The volume of a cone/pyrami with base area A an height h is V = Ah. + ) = + + C. e = e + C. sin = cos + C. = ln + + C ) = ) = + + ln + C Copyright c by The Regents of the University of California

3 Problems Calculus AB Stuents Shoul Know: Solutions... + ) = + ) = + + C + ) = + + ) = C + ) = + ) + C. ivie + = + + ) = + ln + C = + ln ) + C. substitution u =, u = = = ) sin u + C = sin + C u u =. substitution u =, u = + = ) + ) = u + ) u = tan u + C = tan + C +. + = + + ) + = ln + + tan + C = ln + + tan + C For the first integral, substitution u = +, u = + = u u = ln u + C = ln + + C. substitution u =, u = e = eu u = eu + C = + C e. substitution u = ln, u = ln = u = ln u + C = ln ln + C u. substitution u = sin, u = cos sin cos = u u = u + C = sin + C. substitution u = cos, u = sin sin cos = u u = u + C = cos + C = sec + C. substitution u = cos, u = sin sin tan = cos = u = ln u + C = u ln cos + C. = ] = =. integral of even function over symmetric interval = = ] = ) =. integral of o function over symmetric interval. π sin = cos ] π = = cos π cos ) = )+ =. The integral is the signe area, positive for the trapezoi from = to =, negative for the triangle from = to =. f) = + )) )) = =. Copyright c by The Regents of the University of California

4 Problems Calculus AB Stuents Shoul Know: Solutions. a) left enpoints ) + ) + ) ) = b) mipoint rule ) + ) + ) ) = c) trapezoi rule ) ) + ) + ) + ) )) = ) inscribe rectangles ) + ) + ) ) =. intersections = or = ) = at = an =, A = ) = ] =. forwar v = t > so t > istance t ) t = t t ] = backwar v = t < so t < istance t ) t = t t ] = total istance + =.. V = π r r r ) = π π r ] r = πr r ) = πr. V = π + ) r r ) =. π π cos ) = sin π ]π = π +) = π. t t = t] =. vt) = t + ) t = t + t + C, v) = C = vt) = t + t t) = t + t ) t = t + t t + C, ) = C = t) = t + t t +. y =, = y, V = π y) y Copyright c by The Regents of the University of California

5 Problems Calculus AB Stuents Shoul Know: Solutions. V =. y = y ). separation of variables y = y y = y ln y = + C plug in =, y = gives ln = C ln y = + ln y = e +ln = e e ln y = e. =. ) = ) ) = f) = = = + + f) = = f) = f) = = = + + f) = = f) oes not eist, isjunction + = + =. sin =. + = + = oes not eist, isjunction. ± =. e = +. e =. e =. e = +. ln = +. ln = ± + power) + +. ± + +. ± + + power) = because numerator has higher egree = by looking at terms of highest power = because enominator has higher egree Copyright c by The Regents of the University of California

6 Problems Calculus AB Stuents Shoul Know: Solutions. efinition of erivative of f) = e, f ) = e e e = f ) = e. efinition of erivative of f) = e, f ) = e e +h e = f ) = e h h. f) has at least zeros in the open interval, ), with at least zero in, ) an at least zero in, ) because those are where f) goes from positive to negative or vice versa so has to pass through.. Since f) is continuous, it has an absolute minimum value an an absolute maimum value on the close interval [, ]. Specifically, the absolute minimum value is at = an the absolute maimum value is at =.. efinition of efinite integral of f) = + ) + ) = + )] = = or efinition of efinite integral of f) = = ] = =. f f) f ) ) = = =. At = in the open ) interval, ), the instantaneous rate of change is equal to the average rate of change across that interval.. f ) = f) f) or f ) = h f + h) f) h = = h + h) h. f) = ) + ) = + + ) ) vertical asymptotes when enominator is zero = an = f) = =, horizontal asymptotes y = ± + Copyright c by The Regents of the University of California