. Fin erivatives of the following functions: (a) f() = tan ( 2 + ) ( ) 2 (b) f() = ln 2 + (c) f() = sin() Solution: Math 80, Eam 2, Fall 202 Problem Solution (a) The erivative is compute using the Chain Rule twice. f () = ( 2 + ) 2 + 2 + = ( 2 + ) + = 2 + 2 2 2 + 2 2 + 2 (2 + ) (b) We begin by rewriting the function using rules of logarithms. ( ) 2 ln = ln(2 ) ln( 2 + ) = ln(2) ln( 2 + ) 2 + Thus, the erivative is f () = ln(2) ln(2 + ) = ln(2) 2 + = ln(2) 2 + 2 (c) To ifferentiate the function we rewrite it as The erivative is then (2 + ) sin() = e ln sin() sin() ln() = e f () = ln() esin() = e sin() ln() sin() ln() = e sin() ln() = sin() [ cos() ln() + sin() ] [ cos() ln() + sin() ]
Math 80, Eam 2, Fall 202 Problem 2 Solution 2. Use implicit ifferentiation to fin y where tan(2 y) = 4y. Solution: Taking the erivative of both sies of the equation we obtain sec 2 ( 2 y) tan(2 y) = 4y 2 y = 2 y sec 2 ( 2 y) ( 2y + 2 y ) = 2 y 2y sec 2 ( 2 y) + 2 sec 2 ( 2 y) y = 2y Solving for y we obtain y [ 2 2 sec 2 ( 2 y) ] = 2y sec 2 ( 2 y) y = 2y sec2 ( 2 y) 2 2 sec 2 ( 2 y)
Math 80, Eam 2, Fall 202 Problem 3 Solution 3. Consier the function f() = 2 3 + 3 2 2 + 5. (a) Fin the intervals on which f is increasing or ecreasing. (b) Fin the points of local maimum an local minimum of f. (c) Fin the intervals on which f is concave up or concave own. Solution: The first two erivatives of f are: f () = 6 2 + 6 2, f () = 2 + 6 (a) To etermine the intervals of monotonicity of f we begin by fining the points where f = 0. f () = 0 6 2 + 6 2 = 0 6( 2 + 2) = 0 6( + 2)( ) = 0 = 2 or = We summarize the esire information about f by way of the following table: Interval Test #, c f (c) Sign of f (c) Conclusion (, 2) 3 24 + increasing ( 2, ) 0 2 ecreasing (, ) 2 24 + increasing Thus, f is increasing on (, 2) (, ) an ecreasing on ( 2, ). (b) By the First Derivative Test, f has a local maimum at = 2 (f changes sign from + to across = 2) an a local minimum at = (f changes sign from to + across = ). (c) To etermine the intervals of concavity of f we begin by fining the points where f = 0. f () = 0 2 + 6 = 0 = 2 We summarize the esire information about f by way of the following table:
Interval Test #, c f (c) Sign of f (c) Conclusion (, ) 6 concave own 2 (, ) 0 6 + concave up 2 2
Math 80, Eam 2, Fall 202 Problem 4 Solution 4. The prouct of two positive numbers is 200. Fin the two numbers if the sum of their squares is to be as small as possible. Solution: Let an y be the numbers of interest. Since their prouct is known to be 200 we have y = 200 The function to be minimize is the sum of their squares, i.e. 2 + y 2. Solving the constraint equation for y yiels: y = 200 The function is then ( ) 2 200 f() = 2 + = 2 + 2002 2 Note that the omain of f is (0, ) since must be positive. The rest of our analysis begins with fining the critical points of f. f () = 0 2 2 2002 3 = 0 2 4 2 200 2 = 0 4 = 200 2 2 = 200 = 0 2 Since f is concave up on its omain (f = 2 + 6 200 2 / 4 > 0 for all > 0), we know that f has an absolute minimum at = 0 2. The corresponing y-value is y = 200 0 2 = 0 2
Math 80, Eam 2, Fall 202 Problem 5 Solution 5. Let f be a function such that f(3) = an f (3) = 2. (a) Use the linear approimation of f about = 3 to estimate f(3.). (b) Let g be the inverse of f. Fin g() an g (). Solution: (a) The linear approimation of f about = 3 is The approimate value of f(3.) is then L() = f(3) + f (3)( 3) = + 2( 3) f(3.) L(3.) = + 2(3. 3) =.2 (b) By the property of inverses, we know that if f(a) = b then g(b) = a if g is the inverse of f. Thus, since f(3) = we know that g() = 3. Furthermore, the erivative g (b) has the formula g (b) = f (g(b)) Therefore, the erivative g () is g () = f (g()) = f (3) = 2