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Home , Differential calculus, Differentiation rules, Indeterminate form, Vertical tangent

Differential – Definitions, Rules and Theorems Sarah Brewer, Alabama School of Math and Science

Precalculus Review

Functions, Domain and Range : a from to assigns to each a unique the domain of is the set - the set of all real numbers for which the function is defined 푓 푋 → 푌 푓 푋 푌 푥 ∈ 푋 푦 ∈ 푌 is the image of under , denoted ( ) 푓 푋 the range of is the subset of consisting of all images of numbers in 푌 푥 푓 푓 푥 is the independent variable, is the dependent variable 푓 푌 푋 is one-to-one if to each -value in the range there corresponds exactly one -value in the domain 푥 푦 is onto if its range consists of all of 푓 푦 푥 푓Implicit v. Explicit Form 푌 3 + 4 = 8 implicitly defines in terms of = + 2 explicit form 푥 푦 푦 푥 3 푦Graphs− 4 o푥f Functions A function must pass the vertical line test A one-to-one function must also pass the horizontal line test Given the graph of a basic function = ( ), the graph of the transformed function, = ( + ) + = + + 푦 푓 푥, can be found using the following rules: 푐 - vertical stretch (mult. -values by ) 푦 푎푓 푏푥 푐 푑 푎푓 �푏 �푥 푏�� 푑 - horizontal stretch (divide -values by ) 푎 푦 푎 – horizontal shift (shift left if > 0, right if < 0) 푏푐 푥 푐 푏 푐 > 0 < 0 푏 - vertical shift (shift up if 푏 , down if 푏 ) You푑 should know the basic 푑graphs of: 푑 = , = , = , = | |, = , = , = , 3 1 = sin , =2 cos , 3 = tan , = csc , = sec , = cot 푦 푥 푦 푥 푦 푥 푦 푥 푦 √푥 푦 √푥 푦 푥 For푦 further푥 discussion푦 푥 of 푦graphing,푥 see푦 Precal 푥and푦 Trig Guides푥 푦 to Graphing푥

Elementary Functions Algebraic functions (, radical, rational) - functions that can be expressed as a finite number of sums, differences, multiples, quotients, & radicals involving Functions that are not algebraic are trancendental (eg. trigonometric,푛 exponential, and logarithmic 푥 functions)

Polynomial Functions ( ) = + +. . . + + + = leading푛 coefficient푛− 1 2 푓 푥 푎푛푥 푎푛−1푥 푎2푥 푎1푥 푎0 = lead term 푎푛 = 푛degree of polynomial (largest exponent) 푎푛푥 = term (term with no ) 푛 푎Lead0 for end behavior: even푥 functions behave like = , odd functions behave like = 2 3 Odd and Even Functions 푦 푥 푦 푥 = ( ) is even if ( ) = ( ) = ( ) is odd if ( ) = ( ) 푦 푓 푥 푓 −푥 푓 푥 푦Rational푓 푥 Functions푓 −푥 −푓 푥 ( ) ( ) = , , , ( ) 0 ( ) 푝 푥 푓 푥 푝 푞 − 푝표푙푦푛표푚푖푎푙푠 푞 푥 ≠ Zeros of푞 a 푥function Solutions to the equation ( ) = 0; When written in the form ( , 0) are called the x-intercepts of the function 푓 푥 Y-intercepts 푥 Found by evaluating (0) (plugging 0 in for x)

Vertical 푓 Found by setting the denominator equal to 0 and solving for x. Note that any factors found in both numerator and denominator will result in holes rather than vertical asymptotes. The graph will approach but never cross a vertical .

Horizontal/Oblique Asymptotes If degree of the numerator is smaller than the degree of denominator, there will be a horizontal asymptote at = 0. If the degree of the numerator is the same as the degree of the denominator, there will be a horizontal 푦 asymptote at = , where is the constant ratio of the leading coefficients. If the degree of the numerator is one higher than the degree of the denominator, there will be an 푦 푐 푐 oblique asymptote (one degree higher - linear, 2 degrees higher, quadratic, etc), found by long division.

Limits

Informal Definition of a If ( ) becomes arbitrarily close to a single number as approaches from either side, the limit of ( ), as approaches , is . 푓 푥 퐿 푥 푐 푓Note:푥 the푥 existence or nonexistence푐 퐿 of ( ) at = has no bearing on the existence of the limit of ( ) as approaches . 푓 푥 푥 푐 Formal푓 푥 Definition푥 of a푐 Limit Let be a function defined on an open containing (except possibly at ) and let be a . The statement lim ( ) = means that for each > 0, there exists a > 0 such that if 푓 푐 푐 퐿 0 < | | < , then | ( ) | < . 푥→푐 푓 푥 퐿 휀 훿 Translation:푥 − 푐 If given훿 any푓 arbitrarily푥 − 퐿 small휀 positive number , there exists another small positive number such that ( ) is -close to whenever is -close to , then the limit of as approaches exists 휀 훿 and is equal to . 푓 푥 휀 퐿 푥 훿 푐 푓 푥 푐 Note: if the limit퐿 of a function exists, then it is unique.

Example proof of a limit using the definition Problem: Prove that lim 2 + 5 = 1 휺 − 휹 Proof: We want to show that for each > 0 there exists a > 0 such that |(2 + 5) ( 1)| < 푥→−3 푥 − whenever | ( 3)| < . Since we can rewrite |2 + 5 + 1| = |2 + 6| = 2| + 3|, if we take = , 휀 훿 푥 − − 휀 휀 | ( 3)| < |(2 + 5) ( 1)| = 2| + 3| < 2 = then whenever푥 − − 훿 , we have 푥 푥 푥 . Since we have훿 2 휀 found a that works for any , we have proved that the limit of the function is 1. 푥 − − 훿 푥 − − 푥 ∙ 2 휀 Evaluating훿 Limits Analytically휀 − Basic Limits , , > 0 , , , lim ( ) = , lim ( ) = 1. Constant lim = 퐿푒푡 푏 푐 ∈ ℝ 푛 푎푛 푖푛푡푒푔푒푟 푓 푔 − 푓푢푛푐푡푖표푛푠 푥→푐 푓 푥 퐿 푥→푐 푔 푥 퐾 2. Identity lim = 푥→푐 푏 푏 3. Polynomial lim = 푥→푐 푥 푐 4. Scalar Multiple lim [ 푛( )] 푛= 푥→푐 푥 푐 5. Sum or Difference lim [ ( ) ± ( )] = ± 푥→푐 푏푓 푥 푏퐿 lim [ ( ) ( )] = 6. Product 푥→푐 푓 (푥) 푔 푥 퐿 퐾 7. Quotient lim푥→푐 = , 0 푓 (푥) 푔 푥 퐿퐾 푓 푥 퐿 [ ] 8. Power lim푥→푐 �푔(푥 )� =퐾 퐾 ≠ 푛 푛 푥→푐 Note: If substitution yields , an푓 indeterminate푥 퐿 form, the expression must be rewritten in order to 0 evaluate the limit. 0

Examples: ( )( ) lim = lim = lim (2 3) = 5 1. 2 2푥 −푥−3 푥+1 2푥−3 2. lim푥→−1 푥+1 = lim 푥→−1 푥+1 =푥→lim−1 푥 − − = lim = √2+푥−√2 √2+푥−√2 √2+푥+√2 2+푥−2 1 1 ( ) ( ) lim푥→0 푥 = lim푥→0 푥 ∙ √2+푥+=√2lim 푥→0 푥�√2+=푥+lim√2� 푥→0 √=2+lim푥+√(22 +2√2) = 2 3. 2 2 2 2 2 2 푥+ℎ −푥 푥 +2푥ℎ+ℎ −푥 2푥ℎ+ℎ ℎ 2푥+ℎ ℎ→0 ℎ ℎ→0 ℎ ℎ→0 ℎ ℎ→0 ℎ Squeeze Theorem ℎ→0 푥 ℎ 푥 If ( ) ( ) ( ), that is, a function is bounded above and below by two other functions, and lim ( ) = lim ( ) = , that is, the two upper and lower functions have the same limit, then 푔 푥 ≤ 푓 푥 ≤ ℎ 푥 lim ( ) = , that is, the limit of the function that is “squeezed” between the two functions with 푥→푐 푔 푥 푥→푐 ℎ 푥 퐿 equal limits must have the same limit. 푥→푐 푓 푥 퐿 Example: Prove that lim sin = 0 using the Squeeze Theorem. Solution: 1 sin 1. Multiplying2 both sides of each inequality by yields sin . 푥→0 푥 푥 Now, applying limits, we have lim lim sin lim 2 0 2lim 2 sin 2 0 − ≤ 푥 ≤ 푥 −푥 ≤ 푥 푥 ≤ 푥 Since our limit is bounded above and below2 by 0, by the2 Squeeze theorem,2 we have that 2 푥→0 −푥 ≥ 푥→0 푥 푥 ≥ 푥→0 푥 ↔ ≥ 푥→0 푥 푥 ≥ lim sin = 0. 2 Two푥 →important0 푥 푥 limits that can be proven using the Squeeze Theorem that we will use to find many other limits are: sin 1 cos lim = 1 lim = 0 푥 − 푥 푥→0 푎푛푑 푥→0 Examples: 푥 푥 1. lim = lim = 1 1 = sin 2푥 sin 2푥 3푥 2 2 2 ( ) ( ) 2. lim푥→0 sin 3푥 푥→=0 lim2푥 ∙ sin 3푥 ∙ 3 ∙ ∙ 3 3 = lim = sin 푥+ℎ −sin 푥 sin 푥 cos ℎ+cos 푥 sin ℎ−sin 푥 sin 푥 cos ℎ−1 +cos 푥 sin ℎ limℎ→0 ( sinℎ ) +ℎ→0lim(cos ) ℎ = ( sin ) ℎ0→+0 cos 1 =ℎ cos 1−cos ℎ sin ℎ ℎ→0 � − 푥 ∙ ℎ � �ℎ→0 푥 ∙ ℎ � − 푥 ∙ 푥 ∙ 푥

Continuity

If lim ( ) = ( ), we say that the function is continuous at .

Discontinuities푥→푐 푓 푥 occur푓 푐 when either 푓 푐 1. ( ) is , 2. lim ( ) does not exist, or 푓 푥 3. lim ( ) ( ) 푥→푐 푓 푥 If a discontinuity푥→푐 푓 푥 ≠ can푓 푐 be removed by inserting a single point, it is called removable. Otherwise, it is nonremovable (e.g. verticals asymptotes and jump discontinuities) One-Sided Limits lim ( ) =

lim+ ( ) = 푥→푐 푓 푥 퐿 푙푖푚푖푡 푓푟표푚 푡ℎ푒 푟푖푔ℎ푡 − Examples푥→푐 푓 푥 : 퐿 푙푖푚푖푡 푓푟표푚 푡ℎ푒 푙푒푓푡 | | | | 1, 2 lim = 1 = 1. Note that 1, < 2 푥−2 푥−2 + , 1 푥 ≥ lim푥→2 푥(−2) , ( ) = 푥−2 � 2. 1 , >− 1 푥Since we want the limit from the right, we look at the + 푥 푥 ≤ piece푥→ where1 푓 푥 the푤 functionℎ푒푟푒 푓 푥 is defined� > 1. Plugging 1 into 1 gives us that the limit is 0 . − 푥 푥 Continuity at a point − 푥 A function is continuous at if the following 3 conditions are met: 1. ( ) is defined 푓 푐 2. Limit of ( ) exists when approaches 푓 푐 3. Limit of ( ) when approaches is equal to ( ) 푓 푥 푥 푐 Continuity푓 on푥 an open푥 interval 푐 푓 푐 A function is continuous on an open interval if it is continuous at each point in the interval. A function that is continuous on the entire real line ( , )is everywhere continuous.

Continuity on a closed interval −∞ ∞ A function f is continuous on the closed interval [ , ] if it is continuous on the open interval l ( , ) and lim ( ) = ( ) and lim ( ) = ( ). 푎 푏 푎 푏 + − Examples푥→푎 푓: 푥 푓 푎 푥→푏 푓 푥 푓 푏 1. Discuss the continuity of the function ( ) = on the closed interval [ 1,2]. We know that this 1 function has vertical asymptotes at = 2 and 2= 2, has a horizontal asymptote at = 0, and has no 푓 푥 푥 −4 − zeros. The only discontinuities are at 2 and 2, and both are non-removable. So while we can’t say that 푥 − 푥 푥 the function is continuous on [ 1,2] (because we would need the limit from the left to exist at = 2), − we can say that the function is continuous on [ 1,2). − 푥 2. Discuss the continuity of the function ( ) = . Factoring yields ( ) = . Since the − ( )( ) 푥−1 푥−1 2 1 factors cancel, this tells us that there푓 푥 is a hole푥 +푥 in−2 the graph at = 1,푥 and that푥−1 the푥+ function2 behaves like ( ) = everywhere except at = 1. This latter function has a vertical asymptote at 푥 − 푥 1 = 2. Thus, our original function has a removable discontinuity at = 1 and a non-removable 푓 푥 푥+2 푥 discontinuity at = 2. 푥 − 푥 Continuity of a Composite푥 − Function If is continuous at and is continuous at ( ), then ( )( ) = ( ) is continuous at .

Intermedi푔 ate Value Theorem푐 푓 푔 푐 푓 ∘ 푔 푥 푓�푔 푥 � 푐 If is continuous on the closed interval [ , ] and is any number between ( ) and ( ), then there is at least one number in [ , ] such that ( ) = . 푓 푎 푏 푘 푓 푎 푓 푏 푐 푎 푏 푓 푐 푘 Infinite Limits

means the function increases or decreases without bound; i.e. the graph of the function approaches a vertical asymptote

Finding Vertical Asymptotes x-values at which a function is undefined result in either holes in the graph or vertical asymptotes. Holes result when a function can be rewritten so that the factor which yields the discontinuity cancels. Factors that can’t cancel yield vertical asymptotes.

Example:

has vertical asymptotes at and

has a vertical asymptote at and a hole at

Rules involving infinite limits

Let and 1. [ ]

2. [ ] {

3.

The

The of the line to the graph of at the point is given by

The derivative of f at x is given by

An alternate form of the derivative that we often use to show that certain continuous functions are not differentiable at all points (either having sharp points or vertical tangent lines) is given by

Basic

1. The derivative of a constant function is zero, i.e., for [ ]

2. for [ ]

Special case: [ ]

3. Constant Multiple Rule [ ]

4. Sum & Difference Rules [ ]

Derivatives of Trig Functions 1. [sin ] = cos 푑 2. 푑푥 [cos푥 ] = sin푥 푑 3. 푑푥 [tan 푥] = sec− 푥 푑 2 4. 푑푥 [cot 푥] = csc푥 푑 2 5. 푑푥 [sec푥] = −sec tan푥 푑 6. 푑푥 [csc 푥] = csc푥 cot푥 푑 Rates푑푥 of Change푥 − 푥 푥 ( ) ( ) The average rate of change of a function with respect to a variable is given by = ∆푓 푓 푡2 −푓 푡1 The instantaneous rate of change of a function푓 at an instance of the푡 variable is∆ given푡 by푡2 −푡’(1 ) The rate of change of position is . 푓 푡 푓 푡 The rate of change of velocity is . The position of a free-falling object under the influence of gravity is given by ( ) = + + , 1 2 where = initial height of the object, = initial velocity of the object, =푠 푡32 2 푔푡 9.푣80 푡 / 푠표 푓푡 2 표 0 2 Product푠 Rule 푣 푔 − 푠 표푟 − 푚 푠

[ ( ) ( )] = ( ) ( ) + ( ) ( ) 푑 ′ ′ 푓 푥 푔 푥 푓 푥 푔 푥 푓 푥 푔 푥 푑푥Quotient Rule ( ) ( ) ( ) ( ) ( ) = ( ) ′ ( ) ′ 푑 푓 푥 푔 푥 푓 푥 − 푓 푥 푔 푥 “low� dee �high less high dee2 low, draw the line and square below” 푑푥 푔 푥 푔 푥

[ ( ( ))] = ( ) ( ) 푑 ′ ′ When푓 푔one푥 or more푓 � functions푔 푥 �푔 푥are composed, you take the derivative of the outermost function first, keeping푑푥 its inner function, and then multiply that entire thing by the derivative of the inner function. If more than two functions are composed, you just repeat this process for all inner functions, e.g.

[ ( ( ( )))] = ( ) ( ) ( ) 푑 ′ ′ ′ [푓 푔 ℎ 푥 ]푓 �푔�ℎ 푥 �� 푔 �ℎ 푥 �ℎ 푥 푑푥 ( ( ( ( )))) = ( ) ( ) ( ) ( ) 푑 ′ ′ ′ ′ 푓 푔 ℎ 푘 푥 푓 �푔 �ℎ�푘 푥 ��� 푔 �ℎ�푘 푥 �� ℎ �푘 푥 �푘 푥 푑푥

Example: ( ) = sin 3 sec (5 7 ) + cos First we want to rewrite2 4any roots as fractional powers, and any ( ) as [ ( )] . 푓 푥 �� 푥 − 푥 푥� ( ) = sin (3 [sec(5 7 )] + cos ) 푛 푛 푓 푥 푓 푥 Let's identify our sequence4 of nested2 functions1⁄2 to determine what order we will take . 푓 푥 � 푥 − 푥 푥 � sin , = , = 3 + cos , = sec , = 5 7 The derivatives 1of⁄2 these functions2 are, respectively, 4 푎 푎 푏 1 푏 푐 푥 푐 푑 푑 푥 − 푥 (cos ) , = , = (6 ) sin , = (sec tan ) , = 20 7 2 ′ ′ −1⁄2 ′ ′ ′ ′ ′ ′ 3 So, using푎 푎 the푎 chain� rule,푏 the� derivative푏 푏 of our푐 original푐 − function푥 푐 is 푑 푑 푑 푑 푥 − 1 ( ) = cos 3 sec (5 7 ) + cos (3 [sec(5 7 )] + cos ) 2 2 4 4 2 −1⁄2 푓′ 푥 � �� ([6sec(5푥 − 7푥 )] (sec푥(�5� ∙ � 7 ) tan(5푥 − 7푥 )) (20 푥 7)� ∙ sin ) 4 4 4 3 Implicit Differentiation∙ 푥 − 푥 ∙ 푥 − 푥 푥 − 푥 ∙ 푥 − − 푥 When taking the derivative of a function with respect to , the only variable whose derivative is 1 is , since all other variables are treated as functions of (except variables standing for constants, like ). 푥 푥 So, when we have an equation in and written implicitly (i.e. not solved for y in terms of x) and we 푥 휋 want to find , we take the derivative of both sides, but every derivative involving must treat y as a 푥 푦 function of , meaning it has to be multiplied by the derivative of the "inside" function, . 푦′ 푦 Example: 푥 푦′ + = 3 Taking2 3the derivative of both sides with respect to x, we have 푥 푦 푥푦 2 + 3 = 3 + 3 (we used the power rule twice on the LHS, and the on the RHS) Then we 2put all our′ -terms on one side, factor out the , and divide. 푥 푦 푦′ 푥푦 푦 3 3 = 3 2 푦′ 푦′ (23 ′ 3 )′ = 3 2 푦 푦 − 푥푦 푦 − 푥 ′ 23 2 푦 =푦 − 푥 푦 − 푥 3 3 ′ 푦 − 푥 푦 2 If we want푦 −to find푥 in terms of and , we just take the derivative of , again remembering to treat as a function of . We'll end up with an expression in , , and , so we substitute the expression we 푦′′ 푥 푦 푦′ 푦 found for in terms of just and . Using the above implicit differentiation for , we have 푥 푥 푦 푦′ (3 3 )(3 2) (3 2 )(6 3) = 푦′ 푥 푦 푦′ 2 (3 3 ) ′ ′′ 푦 − 푥 푦′ − − 푦 −3 푥 2 푦푦 − 3 2 푦 (3 32 ) 3 2 2 (3 2 ) 6 3 푦 − 푥 3 3 3 3 = 2 푦 − 푥 푦 − 푥 푦 − 푥 � � 2 �(3− � 3푦 )− 푥 � 푦 � 2 � − � 푦 − 푥 2 2 푦 − 푥 푦 − 푥

Related Rates Related rate problems are basically just applied implicit differentiation problems, typically dealing with 2 or more variables that are each functions of an additional variable. For example, a problem dealing with the volume of a cone, where volume, radius, and height are all functions of . We take the derivative of both sides of our equation implicitly, remembering that each variable is a function of time.

Basic equations/formulas to know: Volume of a Sphere = 4 Surface Area of a Sphere = 4 3 푉 3 휋푟 Volume of a cone = 2 퐴 휋푟 1 Volume of a right prism = ( 2 ) ( ) 푉 3 휋푟 ℎ Example problem: A conical tank푉 is 10 푎푟푒푎feet across표푓 푏푎푠푒 at the∙ 푝푒푟푝푒푛푑푖푐푢푙푎푟top and 10 feet deep.ℎ푒푖푔ℎ 푡If it is being filled with water at a rate of 5 cubic feet per minute, find the rate of change of the depth of the water when it is 3 feet deep.

Steps to solve: 1. Identify knowns/unknowns = 5 = 10 = = = 푟 5 1 ℎ 푟 = 5푤 ℎ푒푛/ ℎ → ℎ 10 2 → 푟 2 푑푣 3 푑푡 = ? 푓푡 푚푖푛 = 3 푑ℎ 2. Identify equation 푑푡 푤ℎ푒푛 ℎ 푓푡 = 1 Note that2 this equation involves variables V, r, and h, all of which are functions of t. When we 푉 3 휋푟 ℎ take the derivative, we will end up with , , . We know the value of , and we're 푑푣 푑ℎ 푑푟 푑푣 trying to solve for , but we don't know푑푡 anything푑푡 푎푛푑 about푑푡 , so we want to rewrite푑푡 the equation 푑ℎ 푑푟 so that it doesn't include푑푡 r at all. To do this, we use the ratio푑푡 of radius to height of the tank. In other problems, this step may include rearranging the Pythagorean theorem, or using trig functions - whatever information you know about the given shape/area/volume. 3. Rewrite equation = = = 1 1 ℎ 2 1 1 4. Take the derivative2 of both sides with respect3 to t (or whatever the independent variable is). 푉 3 휋푟 ℎ 3 휋 �2� ℎ 3 �4� 휋ℎ = 푑푉 1 푑ℎ 5. Rearrange to solve2 for the unknown. 푑푡 4 휋ℎ ∙ 푑푡

= 푑푉 푑ℎ 4 푑푡 6. Plug in known2 values. Note that we are not plugging in any known values until the VERY end of the 푑푡 휋ℎ problem. Otherwise, our derivative would have been inaccurate. ( ) = = / 푑ℎ 4 5 20 2 푑푡 휋3 9휋 푓푡 푚푖푛 Relative Extrema, Increasing & Decreasing, and the First Recall that the derivative of a function at a point is the slope of the tangent line at that point. When the derivative is zero, the slope is zero, when the derivative is positive, the slope is positive, and when the derivative is negative, the slope is negative.

( ) ( ) ( ) = 0 f is either constant or has a relative maximum or minumum 푓′′(푥) > 0 f푓 is푥 increasing 푓 (푥) < 0 f is decreasing 푓′ 푥 Absolute푓′ 푥 Extrema on a Closed Interval To find absolute on a closed interval (the highest and lowest y-values a function achieves on the interval), first take the derivative of the function, set it equal to zero, and solve for x. These values are called critical values or critical numbers. Then plug any of these x-values which lie within the given closed interval, and the x-values of the endpoints of the interval into the original function in order to find the y-values of these points. The largest y-value is your absolute maximum and the smallest y-value is your absolute minimum.

Open Intervals on which a Function is Increasing or Decreasing If we want to discuss the behavior of a function on its entire domain, we take all of the critical numbers we found by setting the derivative equal to zero, along with any other x-values for which the derivative is undefined, and use these numbers to split the number line into intervals. For example, say we have a function with the following derivative: ( 2)( + 5)( 8) ( ) = ( + 1) 푥 − 푥 푥 − The푓′ 푥 critical numbers are2 5, 1, 2, and 8. So, we split up the real number line into the following 푥 intervals, and then choose a number within each interval to plug into the derivative to see if it is positive − − or negative in that interval (if it is positive/negative for one value in that interval, it will be for all values). Note that the denominator is always positive, so we can concentrate on the numerator to determine if f' is positive/negative for each value.

( , 5) ( 5, 1) ( 1,2) (2,8) (8, ) ( 6) < 0 ( 2) > 0 (0) > 0 (3) < 0 (9) > 0 ′−∞ − ′− − ′ − ′ ′ ∞ Thus,푓 the− function f is increasing푓 − on ( 5, 1) 푓( 1,2) (8, ) and푓 decreasing on ( ,푓 5) (2,8).

− − ∪ − ∪ ∞ −∞ − ∪

Concavity and the The second derivative is the rate of change of the slope of a function. When slope increases, a function is concave up. When slope decreases, the function is concave down. When a function changes concavity, we get what are called inflection points. To determine possible inflection points, take the second derivative of the function, set it equal to zero and solve for x. Those values are the x-coordinates of your inflection points. Then, divide up the real number line into intervals based on those numbers, and plug a number from each interval into the second derivative. If positive, the function is concave up on that interval. If negative, the function is concave down on the interval.

( ) ( ) ( ) = 0 f has 푓′′(푥) > 0 f푓 is푥 concave up 푓′′(푥) < 0 f is concave down 푓′′ 푥 The푓′′ 푥 second derivative can also be used to determine whether a particular relative extremum, found using the first derivative, is in fact a maximum or a minimum, without having to determine whether the function is increasing or decreasing on either side of the extremum. If the function is concave down at an extremum, it is a maximum; if the function is concave up at an extremum, it is a minimum.

Optimization Problems When trying to maximize or minimize a function, you're basically using the first derivative test to find relative extrema. When you set the first derivative equal to zero and solve for x, if there is more than one solution, it will usually be obvious that only one of them makes sense as the answer. For example, a negative value for area, radius, etc. certainly doesn't make sense, so you choose the positive value.

These problems will typically involve formulas with multiple variables, and so require multiple equations to solve. Before taking any derivatives, figure out every equation you can that relate the variables, then rewrite the formula you are trying to maximize or minimize by substituting for one or more of the variables so that you end up with a function of a single variable.

Rolle’s Theorem: Let be continuous on the closed interval [ , ] and differentiable on the open interval ( , ). If ( ) = ( ), then there is at least one number in ( , ) such that ’( ) = 0. 푓 푎 푏 Basically, 푎Rolle’s푏 Theorem푓 푎 푓 states푏 that if a continuous, differentiable푐 function푎 푏 achieves the푓 푐 same y-value at both endpoints of a given interval, then there must be some value in that interval there the function has a horizontal tangent line.

Mean Value Theorem: If is continuous on the closed interval [ , ] and differentiable on the open ( ) ( ) ( , ) ( , ) ’( ) = interval , then there푓 exists a number in such that 푎 푏 . 푓 푏 −푓 푎 The Mean푎 Value푏 Theorem is really a generalized푐 푎 version푏 of Rolle’s푓 푐Theorem.푏− 푎It states that if a function is continuous and differentiable on a given interval, then there must be some value in that interval where the slope of the tangent line to the graph of the function is the same as the slope of the through the enpoints of the interval.

Limits at The line = is a horizontal asymptote of the graph of if lim ( ) = or lim ( ) = . If r is a positive rational number and c is any real number, then lim = 0. If is defined when 푦 퐿 푓 푥→−∞ 푓 푥 퐿 푥→∞ 푓 푥 퐿 푐 푟 푟 < 0, then lim = 0. 푥→∞ 푥 푥 푐 푥→−∞ 푟 푥Guidelines for finding푥 limits at infinity of rational functions: If the degree of the numerator is less than the degree of the denominator, then the limit is 0. If the degree of the numerator is equal to the degree of the denominator, then the limit is the ratio of leading coefficients. If the degree of the numerator is larger than the degree of the denominator, then the limit is ± . (the above 3 rules follow the rules for horizontal/oblique asymptotes we learned in Precal) ∞ If the numerator or denominator has a radical, remember that the nth root of x to the n is defined as: , = 푛 | |, 푛 푥 푖푓 푛 푖푠 표푑푑 and√푥 that� absolute value of x is defined as: ,푥 푖푓 푛 푖푠0 푒푣푒푛 | | = , < 0 푥 푖푓 푥 ≥ 푥 � Indeterminate−푥 푖푓 푥Forms and l’Hopital’s Rule 0 0, , 0 , 1 , 0 , and are called indeterminate forms. ∞ 0 l’Hopital’s⁄ ∞⁄∞ Rule∙ ∞: Let and be∞ functions− ∞ that are differentiable on an open interval ( , ) containing , except possibly at itself. Assume that ’( ) 0 for all x in (a,b), except possibly at c itself. If the limit 푓 푔 푎 푏 푐 of f(x)/g(x) as x approaches c produces an 0/0, , ( ) , or ( ), then ( ) 푐 ( ) 푔 푥 ≠ lim = lim ( ) ( ) ∞⁄∞ −∞ ⁄∞ ∞⁄ −∞ 푓 푥 푓′ 푥 푥→푐 푔 푥 푥→푐 푔′ 푥

DERIVATIVE RULES

Power Rule: Product Rule:

[ ] = [ ( ) ( )] = ( ) ( ) + ( ) ( ) 푑 푛 푛−1 푑 ′ ′ 푥 푛푥 푓 푥 푔 푥 푓 푥 푔 푥 푓 푥 푔 푥 푑푥Constant Multiple Rule: 푑푥Quotient Rule: ( ) ( ) ( ) ( ) ( ) [ ( )] = [ ( )] = ( ) ′ ( ) ′ 푑 푑 푑 푓 푥 푓 푥 푔 푥 − 푓 푥 푔 푥 푐푓 푥 푐 푓 푥 � � 2 Sum푑푥 & Difference:푑푥 Chain푑푥 푔 Rule:푥 푔 푥

[ ( ) ± ( )] = ( ) ± ( ) [ ( ( ))] = ( ) ( ) 푑 ′ ′ 푑 ′ ′ 푓 푥 푔 푥 푓 푥 푔 푥 푓 푔 푥 푓 �푔 푥 �푔 푥 푑푥 푑푥 Trig Functions:

[sin ] = cos [tan ] = sec [sec ] = sec tan 푑 푑 2 푑 [ 푥 ] 푥 [ 푥] 푥 [ 푥] 푥 푥 푑푥 cos = sin 푑푥 cot = csc 푑푥 csc = csc cot 푑 푑 2 푑 푥 − 푥 푥 − 푥 푥 − 푥 푥 푑푥 푑푥 푑푥 Exponential and Logarithmic Functions:

[ ] = ln [log ] = ln 푑 푢 푢 푑 푢′ 푎 푎 푢′ 푎 푎 푢 푑푥 푑푥 푢 푎 Inverse Functions: 1 [ ( )] = ( ( )) 푑 −1 푓 푥 ′ −1 푑푥 푓 푓 푥 Inverse Trig Functions:

[arcsin ] = [arccos ] = 1 1 푑 푢′ 푑 −푢′ 푢 2 푢 2 [arctan ] = [arccot ] = 푑푥 √1 +− 푢 푑푥 1√+− 푢 푑 푢′ 푑 −푢′ 2 2 [arcsec 푢] = [arccsc 푢] = 푑푥 | | 푢 1 푑푥 | | 푢 1 푑 푢′ 푑 −푢′ 푢 2 푢 2 푑푥 푢 √푢 − 푑푥 푢 √푢 −