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Final

# MATH 137 Final Review Package.docx

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Mathematics
Course Code
MATH 137
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Frank Zorzitto

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Waterloo SOS MATH 137 Final Exam AID December 10th 2011 By: Arjun Sondhi Fall 2011 D IFFERENTIALC ALCULUS Power Rule. ( ) ( ) Product Rule. , ( ) ( )- ( ) ( ) ( ) ( ) Quotient Rule. [ ( ) ] ( ) ( ) ( ) ( ) ( ) , ( )- Chain Rule. Example. Find the derivative of ( ) ( ) Note that this function can be rewritten as a composition ) , where . Therefore, by Chain Rule, we have: ( ) ( ) The chain rule in this case is referred to as the power of a function rule. It is used when the exponent applies not just to a single x, but to an entire function. Exponential Derivatives. ( ) Note: If e or a are raised to some function (ie. not just x on its own), then we must apply Chain Rule Example. Find the equation of the tangent to ( ) at Using Product Rule and Chain Rule: ( ) , - , - At x = 1: ( ) ( ) ( ) Thus, the slope of the tangent at x = 1 is . At x = 1, we have ( ) . Therefore, in slope- point form, the equation of the tangent is: ( ) Trigonometric Derivatives. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Note: If a trig function is applied to some function (ie. not just x on its own), then we must apply Chain Rule ( ) Example. Find the points where the tangent to ( ) ( )is horizontal. A horizontal tangent has a slope of 0. Using quotient rule: , ( )-( ( )) , ( ) -( ( )) ( ) ( ( )) ( ) ( ) ( ) ( ) ( ( )) By the Pythagorean trig identity: ( ) ( ) ( ( )) Setting ( ) , we have: ( ) ( ) ( ) Implicit Differentiation. We use implicit differentiation on a function when the function in question is not written explicitly in terms of y. Differentiate every term, on both sides of the equation. Recall that we treat y as a function, so when we differentiate a term involving y, we need to apply Chain Rule and multiply by the derivative in Leibniz notation. Then, it is possible to solve explicitly.for Example. Find the derivative of the unit circle Differentiating all terms: ( ) Solving explicitly for the derivative: ( ) Example. Find the derivative of ( ) Differentiating all terms: ( ) ( ) ( ) ( ) Collecting like terms, and solving explicitly for the derivative: ( )( ( )) ( ) Inverse Trigonometric and Logarithmic Derivatives. By using implicit differentiation, we can find the derivatives for more complex functions. ( ( )) √ ( ( )) ( ( ) ) ( ( )) ( ) Note: As before, if one of these functions is applied to some other function (ie. not just x on its own), then we must apply Chain Rule. Example. Differentiate ( ) Applying the definition of logarithmic derivatives and Chain Rule: ( ) Logarithmic Differentiation. When we have a large function that would require several iterations of Product and/or Quotient Rule, applying logarithmic differentiation may help to simplify it. To use logarithmic differentiation, take the ln of both sides of the equation. Then, simplify using log laws. Once this is done, implicitly differentiate. √ ( ) Example. Find the derivative of ( ) Taking the ln of both sides: √ ( ) ( ) ( ( ) ) Simplifying with log laws: ( ) (√ ( ) ) ( ( )) ( ) (√ ) (( ) ) ( ) ( ( ) ) ( ) ( ) ( ) ( ) ( ( )) ( ) ( ) ( ) ( ( ) ) Applying implicit differentiation: ( ) ( ) ( ) , ( )- ( ) √ ( ) ( ) , - ( ) ( ) Example. Find the derivative of Method 1: Logarithmic Differentiation ( ) ( ) ( ) ( ) Applying implicit differentiation: , -( ( )) ( ) ( ( ) ) ( ( ) ) Method 2 Note that ( ) , where blah can be anything. ( ) ( ) Differentiating both sides: ( ( )) ( ) (, -( ( )) ( ) ) ( ( ) ) A PPLICATIONS OF D IFFERENTIALC ALCULUS Limits Revisited. Sometimes, it can be easier to take a derivative than a limit. If we get a limit expression that mimics the form of the first principles definition of the derivative, then instead of taking that limit, we can just evaluate the derivative. ( ) Example. Evaluate Since ( ) : ( ) ( ) ( ) Replace x with h, where a = 1 and ( ) ( ): ( ) ( ) ( ) ( ) ( ) Therefore: ( ) Through this method, we can derive the following two definitions of in the form of limits: ( ) ( ) Related Rate Problems. A related rate problem is a question involving applied differential calculus. We interpret the derivative as a rate of change. Usually, there is more than 1 equation and more than 1 unknown variable and/or rate to deal with. We use the relation between rates and variables to find the answer. Example. Two runners are running perpendicular to each other. Runner A runs at a rate of 5 m/s, while Runner B runs at a rate of 4 m/s. After 10 seconds, how quickly is the distance between the runners increasing? Step 1. Define Variables Let x be the distance travelled by Runner A. Let y be the distance travelled by Runner B. Let z be the distance between the runners. Let t be the time in seconds since the runners started running. *IMPORTANT: y is NOT a function of x! Step 2. Information Given Step 3. Information Needed Step 4. Solve Applying implicit differentiation: ( ) ( ) ( ) ( ) ( ) ( ) At t = 10, . √ ( ) ( ) √ √ ( ) ( ) ( √ ) √ √ Thus, our answer i√ m/s. Extreme Values. A function f on domain D has an absolute maximum at c if ( ) ( ) or an absolute minimum at c if ( ) ( ) . A function f has a local maximum or minimum at c if ( ) ( ) ( ) ( ) in a small open interval around c. A function f has a critical number at c if ( ) or ( ) does not exist. Fermat’s Theorem. If f has a local maximum or minimum at c and ( ) exists, then ( ) Logical Equivalent. If f has a local maximum or minimum at c, then ( ) OR ( ) does not exist. Note that the converse is not true. For example, for ( ) , ( ) , but there is no local max/min at 0. Extreme Value Theorem. If f is continuous on [a, b] then f attains an absolute maximum value ( ) and an absolute minimum value ( ) for some , -. If f satisfies the conditions of the EVT, we use the Closed Interval Method (aka Max/Min Algorithm) to find the maximum and minimum: 1. Find all critical values in (a, b) 2. Evaluate f at all critical values and at the endpoints 3. The largest of these is the absolute maximum, the smallest is the absolute minimum Example. Find the maximum and minimum of ( ) ( )on [1, 3] f is continuous on [1, 3] [ ]( ) ( ), - ( ) ( ) ( ) exists for all , - ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Therefore, the maximum is and the minimum is 0. Mean Value Theorem. If f is a function such that: 1. f is continuous on [a, b] 2. f is differentiable on (a, b), ( ) ( ) ( ) then ( ) such that Special Case Rolle’s Theorem. If f satisfies the third condition that ( ) ( ) (in addition to the MVT conditions), then ( ) such that ( ) Example. Given that ( ) and that ( ) , find the range of possible values for ( ). is differentiable for all is continuous for all Therefore, we can apply the MVT for any interval. Since we know ( ) and want to know ( ), let’s choose (2, 4). ( ) ( ) ( ) By the MVT, ( ) such that ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Constant Function Theorem. If ( ) ( ) then ( ) is constant on ( ). Corollary. If ( ) ( ) ( ), then ( )( ) is constant on ( ). Increasing Function Theorem. If ( ) for all x in an interval A, then ( ) is increasing on A. If ( ) for all x in an interval A, then ( ) is decreasing on A. Example. Show that ( ) ( ) has exactly one root. f is continuous for all x,( ) , and ( ) Thus, by the IVT, f has at least one root. ( ) ( ) for all x Therefore, by the IFT, f is increasing for all x. Thus, f has exactly one root. Curve Sketching. 1. Domain  Look for restrictions 2. Symmetry  Check for even or odd symmetry 3. Asymptotes  Vertical asymptotes for rational functions  Horizontal asymptotes by examining ( ) 4. Intervals of Increase/Decrease  Increasing/Decreasing Test: Use the Increasing Function Theorem to determine the increasing and decreasing intervals. 5. Local Maximum/Minimum  First Derivative Test: If ( ) changes from negative to positive at a critical number c, then there is a local minimum at c. If ( ) changes from positive to negative at a critical number c, then there is a local maximum at c. ( ) does not change sign at c, then there is no local extrema at c. 6. Concavity and Points of Inflection  If ( ) for all x in an interval, then f is concave up on the interval.  If ( ) for all x in an interval, then f is concave down on the interval.  Second Derivative Test: If f is continuous at a point P and it changes from concave up to concave down or vice versa, then P is a point of inflection. 7. Sketch Example. Sketch ( ) ( ) 1. Domain: We can’t take the ln of a number less than or equal to 0, * + 2. Symmetry: Neither even nor odd symmetry 3. Asymptotes: As ( ) Since * +, we need to examine ( ) (by L’Hôpital’s Rule) 4. Intervals of Increase/Decrease ( ) , - ( ) ( )[ ]
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