Lecture for Week 4 Secs 3 1 3 Derivatives Finally we get to the point 1 There are two equivalent forms of the definition of a derivative f x f a f a lim x a x a f x h f x f x lim h 0 h One uses whichever seems most natural or useful in a given context Usually the second is more convenient for calculating derivatives from first principles 2 I assume that you have read all the explanations in the book so I will mostly do examples Exercise 3 1 27 Find the derivative of f x x4 using the definition of derivative Actually we already did this for the special case x 1 remember 3 x h 4 x4 f x lim h 0 h I showed you how to do this in Lecture 2 slides 25 27 x h 4 x4 4x3 h O h2 where O h2 is shorthand for terms containing 4 h2 or even higher powers So 3 2 4x h O h f x lim h 0 h 3 lim 4x O h h 0 3 4x 5 Now I hear some of you saying Why are we doing this Don t we all know that the derivative of xn is nxn 1 And that easy formula is in the next section The immediately practical reason is that you ll have to do something like this on the first exam and maybe also on the final so you d better get used to it The deeper reason is that formulas are use6 less unless you know what they mean In science and engineering courses you ll study how equations involving derivatives arise out of the physics of a problem by considering how some variables change as other variables change There the derivative is the answer not the question In those situations you don t know beforehand what the functions involved are you just have to understand that one is the derivative of another Applying the definition to particular functions is simple less abstract practice for that kind of 7 thinking Also of course we will need the definition and the skills of working with it to find derivatives of nonpolynomial functions such as sin x later So let s do one more pretending that none of you have ever heard of the chain rule and the power rule 8 Exercise 3 1 23 Find the derivative of g x definition of derivative 9 1 2x using the 1 2x 2h 1 2x g x lim h 0 h Somehow we have to squeeze an h out of the numerator so we can cancel it Experience shows that multiplying top and bottom by 1 2x 2h 10 1 2x might help See Lecture 3 slide 14 1 2x 2h 1 2x h 0 h 1 2x 2h 1 2x 2 lim h 0 1 2x 2h 1 2x 1 1 2x g x lim 11 Ok let s go on to the next section and use the power rule and other theorems Exercise 3 2 7 Differentiate Y t 6t 9 12 This is a simple instance of the power rule Y t 9 6t 10 54t 10 54 alias 10 t 13 Exercise 3 2 15 Differentiate y x2 4x 3 x 14 Hard way Use the quotient rule Easy way x x1 2 so y x3 2 4x1 2 3x 1 2 y 23 x1 2 2x 1 2 23 x 3 2 If you like you can write this back in terms of square roots as in the book s answer 15 Exercise 3 2 37 Find the equation of the tangent line to the curve 4 y x x at the point 2 4 16 The slope of the line is the derivative of the function 4 y 1 2 x 1 y 2 1 0 1 So the tangent line is the horizontal line through the point y 4 17 Usually the tangent will not be horizontal so you have to use the point slope form of a line y f a x a f a is the tangent line to the graph of f at the point a f a 18 If g is the derivative of f then f is called an antiderivative of g Later we will see that any two antiderivatives of g differ only by a constant provided that the domain of the functions is the whole real line or a single interval You need to find antiderivatives in firstsemester physics so we will treat them briefly now 19 Exercise 5 7 1 p 353 Find the most general antiderivative of f x 12x2 6x 5 Exercise 5 7 9 Find the most general antiderivative of t3 2t2 g t t 20 f x 12x2 6x 5 Use the power rule in reverse remembering to divide each term by the new exponent F x 4x3 3x2 5x C C arbitrary constant is the most general antiderivative of f As you may know already later we ll write Z 2 3 2 12x 6x 5 dx 4x 3x 5x C 21 t3 2t2 g t t t5 2 2t3 2 So G t 27 t7 2 54 t5 2 C is the most general antiderivative Replacing the square root factor by a power which was convenient in an earlier example slide 15 is almost essential now 22 Exercise 3 3 7 The position function of a particle is s t3 4 5t2 7t for t 0 When does the particle reach a velocity of 5 m s 23 s t3 4 5t2 7t for t 0 The velocity is the derivative of the position v 3t2 9t 7 5 3t2 9t 7 3t2 9t 12 0 0 t2 3t 4 t 4 t 1 The root t 1 is outside the stated domain so t 4 s 24 Exercise 3 3 11 A stone is dropped into a lake creating a circular ripple that travels outward at a speed of 60 cm s Find the rate at which the circle s area is increasing after a 1 s b 3 s c 5 s 25 We need the formula for the area of a circle of radius r A r 2 We also need to know what the radius is at each time from the problem statement that is clearly r 60t Therefore A 60 2 t2 3600 t2 dA 7200 t dt To get the numerical answers insert t 1 3 5 Units cm2 s 26
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