1 2 x Solution. The function f x is only defined when x 0, so we will assume that x 0 for the remainder of the solution. f x. f x h f x.

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1 Problem. Let f x x. Using te definition of te derivative prove tat f x x Solution. Te function f x is only defined wen x 0, so we will assume tat x 0 for te remainder of te solution. By te definition of te derivative, f x f x f x lim 0 lim 0 x x We multiply te numerator and denominator by te conjugate of x x: f x lim 0 lim 0 lim 0 lim 0 x x x x x x x x x x x x x x x x lim 0 x x Since te square root function is continuous, x x as 0, so tat as 0, x x x. Tus if x 0, te limit in te definition of te derivative does not exist. If x 0, te quotient rule for limits gives us tat te limit in te definition of f x is exactly f x x Problem. Given tat sin lim 0 prove, using te definition of te derivative, tat if f x sinx, ten f x cosx. Solution. Let f x sinx. By te definition of te derivative, f x f x f x lim 0 sin x sinx

2 Using te formula sin α β f x sin α cos β sin β cos α, we find sinxcos sincosx sinx lim 0 lim 0 sin cosx cos sinx Since sin as 0, we get (using additivity of limits) tat Using te alf-angle formula we can rewrite tis as f x sin lim cosx lim 0 0 cosx lim 0 cos sin α cosα cos sinx sinx sin f x cosx lim 0 sin cosx lim sin 0 Since 0 as 0, we find tat lim sin 0 lim 0 sin. Tus, using multiplicativity of limits, we can write f x sin cos x lim 0 cosx lim 0 sin Since sin is a continuous function, its limit at 0 is sin0 f x cosx lim sin 0 0. We terefore conclude tat Problem. Find given tat: (a) y 4x ; (b) y sin x! sin x cos x ; (c) y (d) y sec tan x y $. sin x#" cos x#" ; Solution. (a) d 4x 4d x 4 x 8x

3 (b) (c) d sinx sinx cos x % d sinx" (d) We use implicit differentiation: d sinx cosx d $ sinx cos x % d sinx d x cosx cos x (& sinx ' sin x $ cosx & sinx ' d cosx cosx d x cosx cos x 4x cosx sin x sin x 9x cosx 4xcos x cos x 9x sin x sin x d cosx" sinx cos x cosx cosx cosx sinx sinx cosx cos x cosx sin x sinx cosx cosx sinx cosx sec tan x y $ tan x y d x y sec tan x y % tan tan x y $ sec x y sec tan x y % tan tan x y $ sec x y sec tan x y % tan tan x y $ sec x y % Tus, by moving te terms involving ) to te left side of te equation, we get sec tan x y $ tan tan x y $ sec x y % sec tan x y $ tan tan x y % sec x y Solving tis we get sec tan x y $ tan tan x y % sec x y sec tan x y % tan tan x y % sec x y Problem 4. Find a point were te curve y x x x 5 as a orizontal tangent.

4 Solution. Te curve as a orizontal tangent at te points were is zero. We find x 6x x x * x Tis expression is zero wen x +. Te corresponding value of y is 5 4. Terefore, te curve as a orizontal tanget at te point 4. 4 Problem 5. A troug is 0 feet long and its ends ave te sapes of isosceles triangles tat are ft across at te top and ave a eigt of foot. If te troug is filled wit water at a rate of ft ) min, ow fast is te water level rising wen te water is 9 inces deep. Solution. Let V t be te volume of te water in te troug at a given time t, and let t be te eigt of te water level at te same time. Ten V t is te rate at wic te tank is filled, and t is te rate at wic te water level is rising. Tus we are given V t and we need to find t. If te water level is t, ten te amount of water te troug olds is given by its lengt (0ft) times te area of te part of te end wic is covered wit water. Te sape of tis part is an isosceles triangle wit eigt, wic is similar to te given triangle (wose eigt is ft and widt is ft). Tus te desired area is t times te area of te end of te troug, wic is Tus ) V t, 0 Differentiating tis equation implicitely in t we get Solving for t gives V t 5 d t dt t - 5 t 5 t t, 0 t t t, V t 0 t We are given tat at te time we are interested in, V t - and t is 9 inces (i.e., 9) ) 4 of a foot). Tus t, Problem 6. Let C be te curve defined by x xy y and wic goes troug te point. Wat is te slope fo te tangent line to C at? Solution. We implicitely differentiate te quation for C: Tus x y x y x y. x y 0

5 Tus x y x y At te given point x and y. Substituting tese in we get Tus te slope is.. Problem 7. Give an example of a function tat is continuous on I 0/ but not differentiable at at least two points in. Solution. Let f x 4 x 5 x. We claim tat te function is not differenentiable at and, but is continuous. Te function f x is a continuous function, because te functions x and x are continuous, and terefore so is teir product, f x. To see tat f x is not differentiable at, we compute its derivative at tat point using te definition of te derivative: f lim 0 9 lim 0 Since x is a continuous function, lim 0 :; < terefore conclude tat f lim 7 lim 0 0. Using te product rule for limits, we lim 0 But tis limit does not exist. Indeed, if = 0, ten :> and so?? >. If 0,?? Tus lim 0@ + lim 0A Since te one-sided limits are different, te limit does not exist. Tus f does not exist. Note tat Moreover, f x *B7 x 57 x 47 x C9$ x CD f x d f x E f x 5. by te cain rule. Tus if f x exists, ten so must f x. We conclude tat since f does not exist, ten f cannot exist, eiter. Tus f x is not differentiable at eiter of te two points F.

6 Problem 8. A function f x and its first two derivatives are graped below. Label wic one is wic. 6 Solution. Let us label te graps as in te picture and let us call te function f x. Suppose tat te grap of f x is given by (c). Since f x is increasing, its derivative is positive, and so its grap could be eiter (a) or (b). Suppose tat (a) is te grap of f x, so tat te grap of f G x is te remaining coice, (b). Since te function wit grap (a) is decreasing, its derivative f H x must be negative. But tis contradicts te assumption tat te grap of f I x is (b). Similarly, if we assume tat te grap of f x is (b), so tat te grap of f I x is (a), we again arrive at a contradiction, since f x is decresing, so tat f I x J= 0, wic is not te case wit (a). Suppose next tat te grap of f x is given by (a). Since f x is decreasing, its derivative must be negative, so tat te grap of f x is (c). Te slope of te grap (a) at x K 4 appears to be, since te line wit tis slope is tangent to (a). Tus f 4 K. However, 0 we see tat (c) is approximately 4 at x K 4, wic is a contradiction. 0 Suppose lastly 0 tat te grap 0 of f x is given by (b). Since f x is decreasing, its derivative must be (c) and its second derivative must be (a). Tis is te only remaining coice and must terefore be te answer.