MATH 124. Midterm 2 Topics
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1 MATH 124 Midterm 2 Topics Anything you ve learned in class (from lecture and homework) so far is fair game, but here s a list of some main topics since the first midterm that you should be familiar with: 1. Di erentiation Techniques: (a) Chain rule: find derivatives of compositions of functions; d dx f(g(x)) = f 0 (g(x))g 0 (x) (b) Implicit di erentiation: find dy for an expression when we can t explicitly solve for y; dx find the tangent line through a point given an implicitly defined curve; know/know how to find inverse trig derivatives (c) Logarithmic di erentiation: find derivatives of logarithmic functions; find derivatives of exponential functions where the base and the exponent contain a variable 2. Parametric Di erentiation: understand the parametric definition of a curve; be able to find x and y derivatives, interpret them as horizontal and vertical rates of change and use them to find speed, s(t) = p x 0 (t) 2 + y 0 (t) 2 ;beabletofind dy from a parametric definition dx of a curve; be able to find tangent lines to parametric curves. 3. Related rates: know how to set up equations relating given quantities using various geometric and algebraic facts (volume and area formulas, Pythagorean theorem, law of cosines, trig identities, similar triangles, etc); be able to solve problems involving two or more related quantities that are all changing with time. 4. Linear Approximation: know how to approximate function values near a point using tangent line approximation; f(a)+f 0 (a)(x a). 5. Min/Max of functions: know definition of critical points, local and absolute min/max; finding min/max of functions. This package includes a some reminder notes on the main di erentiation techniques and how to approach related rates problems, a collection of practice problems on the above main topics, and one complete old midterm (more available on the archive link on the syllabus). Solutions will be available on my website ( awiebe) sometime later this week. Happy studying! Amy 1
2 Di erentiation Techniques Chain Rule If then F (x) =f(g(x)), F 0 (x) =f 0 (g(x)) g 0 (x). When to use the chain rule: Look for a function within a function Think about working from the outside in: 1. Di erentiate the outermost function, don t change the input: if y = f(u), find f 0 2. Now multiply by the derivative of the input: y 0 = f 0 (u) u 0 3. Repeat until the input is just x: findu 0 using steps 1 and 2 until u = x. Example 1. Find y 0 if y =(cos( p x)) 3. In steps: 1. y = u 3, y 0 =3u 2 u 0 2. u =cos(m), u 0 = sin(m) m 0 3. m = p x, m 0 = 1x 1/2 stop since input is now x 2 4. All together: y 0 =3(cos( p x)) 2 {z } derivative of u 3 p sin( x) {z } derivative of cos m 1 2 x 1/2 {z } derivative of p x 1
3 Implicit Di erentiation We can di erentiate any expression involving x and y, by using the chain rule on terms involving y: d dx g(y) =g0 (y) y 0 Then we can rearrange the equation to solve for y 0. [Sometimes your final expression will have y still in it.] When to use implicit di erentiation: Anytime we want to di erentiate an equation that doesn t look like y = f(x) (ifitisdi culttosolvefory) Use our normal di erentiation rules to take the derivative of each term, and every time you take a derivative of a y term, multiply by y 0. Example 2. Find dy dx when x =0if xy2 + y + x 2 =1. d dx (xy2 + y + x 2 )= d dx (1) di erentiate both sides x d dx (y2 )+y 2 d dx (x)+y0 +2x =0 product rule on x y 2 x(2y y 0 )+y 2 + y 0 +2x =0 y 0 (2xy +1)= 2x y 2 y 0 = 2x y2 2xy +1 Notice when x =0, we have 0y 2 + y +0 2 =1, so y =1and then y 0 = 2(0) 12 2(0)(1) + 1 = 1. Implicit di erentiation can also help us remember inverse trig derivatives: Example 3. If y =arcsin(x), then sin(y) =x cos(y)y 0 =1 y 0 = 1 cos(y) y 0 1 = p 1 sin 2 (y) y 0 1 = p 1 x 2 di erentiate both sides use trig identity: sin 2 +cos 2 =1 use the definition of y : sin(y) =x 2
4 Logarithmic Di erentiation Once we know d dx ln(x) = 1 x instead of di erentiating an equation y = f(x) directlywecan: 1. Take logs of both sides: ln(y) =ln(f(x)) 1 2. Implicitly di erentiate with respect to x (chain rule!): y y0 = 1 f 0 (x) f(x) 3. Solve for y 0 : y 0 1 = y f 0 (x) = f 0 (x) [since we substitute y = f(x)] f(x) Notice the answer we get is exactly the derivative we want y 0 = f 0 (x), but ln(f(x)) might be a simpler expression to di erentiate than f(x). When to use logarithmic di erentiation: Usually when we want to di erentiate an function involving a logarithm or with variables in the base and the exponent: y = f(x) g(x). (Can also use on any expression that logs will simplify.) Example 4. A silly example to show why ln(f(x)) might be easier to di erentiate than f(x): Find y 0 for y =(2x 1) p x 3. y =(2x 1) p x 3 ln(y) =ln(2x 1) + ln( p x 3 ) take ln of both sides Remember: ln turns into +, so we get rid of the product rule step! 1 y y0 = 1 2x 1 2+ p 1 3 x 3 2 x1/2 di erentiate both sides with chain rule 1 y 0 = y 2x 1 2+ p 1 3 x 3 2 x1/2 solve for y 0 y 0 =(2x 1) p 1 x 3 2x 1 2+ p 1 3 x 3 2 x1/2 replace y using original function. Exercise: check this is the same derivative as if you use the product rule! Example 5. An example where we really have to use logarithmic di erentiation ( Do not try to use the power rule here!): Find dy dx if y =(cos(x))x2. y =(cos(x)) x2 ln(y) =x 2 ln(cos(x)) take ln of both sides 3
5 1 y y0 = x 2 d dx 1 y y0 = x 2 1 y 0 = y (ln(cos(x))) + ln(cos(x)) d dx (x2 ) sin(x)+ln(cos(x)) 2x cos(x) x 2 1 sin(x)+ln(cos(x)) 2x cos(x) y 0 =(cos(x)) x2 x 2 1 sin(x)+ln(cos(x)) 2x cos(x) di erentiate both sides solve for y 0 replace y using original function. Another way to use logs to di erentiate is to use the fact that any exponential function can be written as e stuff, and this is an easy function to di erentiate. Example 6. Find y 0 when y =2 3x. Rewrite the equation with base e: 3x ln(2) y = e y 0 = e 3x ln(2) 3ln(2) a b = e b ln(a) (1) di erentiate using chain rule y 0 =2 3x (3 ln(2)) rewrite first factor equation (1) 4
6 Parametric Di erentiation If (x, y) = (x(t), y(t)), then at any point t = a where x0 (a) 6= 0, we have dy dx = t=a y 0 (a). x0 (a) That is, dy dy/dt =. dx dx/dt When to use parametric di erentiation: When you are given a curve defined in terms of a third parameter t If we think about t as time: The parametric equations give our (x, y)-position at time t, x0 (t) is horizontal velocity y 0 (t) is vertical velocity, the actual direction we are moving (the slope of the tangent line) is p our overall speed is x0 (t)2 + y 0 (t)2 dy dx = y 0 (t), x0 (t) Here is a picture to help you remember: Notice if x0 (t) = 0 and y 0 (t) 6= 0, we get a vertical tangent line. Notice if y 0 (t) = 0 and x0 (t) 6= 0, we get a horizontal tangent line. Example 7. Find the equation of the tangent line to the curve (x, y) = (t2 cos(t), 3t point where t =. Find point: Find derivatives: x = 2 cos( ) = 2, x0 (t) = sin(t)t2 + 2t cos(t), dy dx dy dx at Line equation: = y= 3 sin( ) 2 +2 cos(pi) 3 (x + 2 ) = 1 5 y = 3 1 y 0 (t) = 3 1) at the
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20 Related Rates 1. Draw and label a diagram: Label any unchanging quantities with their numbers. If a quantity is changing in the problem, it must get labeled with a variable! 2. Write what you know and what you want: List any rates of change given in terms of your variables. variables they give you in the question. Write down the quantity you want to find! Write down the value of any 3. Write down a relationship involving the quantity you want to find: Use your diagram and algebra/geometry facts such as volumes of objects (cones, spheres, prisms, cylinders, etc.,) areas of shapes (triangles, trapezoids, circles, etc) Pythagorean theorem law of cosines/sines trig identities similar triangle ratios distance formula... etc 4. Di erentiate your equation from above with respect to t: Ask yourself if you know all the values in your di erentiated equation except for the quantity you are trying to find: If YES: go to 5 If NO: decide which quantities you are missing and use your equation from 3 to find them, or go back to step 3 and write down a new relationship involving whatever quantity you are still missing. Repeat until you have all the values you need. 5. Plug in given values: ONLY NOW do we plug in the information from 2. Solve for the quantity we were trying to find. Linear Approximation The linear approximation (or linearization) of a function f(x)atthepointx = a is just the equation of the tangent line: L f (x) =f(a)+f 0 (a)(x a). For points near a, we can estimate the value of the function f using the fact that the tangent line is close to the actual function near the point of tangency: If x = b is close to x = a, then f(b) f(a)+f 0 (a)(b a).
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