Alvin Jin Math 16A: Worksheet 2 9/4/14 Today, we’ll be trying to get to the idea of a derivative. We’ll start with something basic: the slope of a straight line. 1. Find the slope of the line passing through the points (6, −2) and (9, 4). 2. Find the equation of a line passing through (−1, 2) with slope 3. 3. Find the equation of a line perpendicular to y = 21 x + 2 containing the point (1, −3). The properties that we’re working with above apply to the whole line we’re working with. But, what if we want to see what the slope of the line or graph is at a particular point? 4. Try finding the slope of f (x) = x2 at (1, 1). To do this, apply the concept of slopes of straight lines above (we use something called the tangent line). Use the fact that the tangent line to the curve at the point (1, 1) is given by y = 2x − 1. 5. Do the same thing with f (x) = x3 at (1, 1). Note that the tangent line to (1, 1) for this function is given by y = 3x − 2. Alright, so now that you’ve seen the concept of instantaneous slope (that is, slope of a curve at a point), how do you find the equation of the tangent line without me giving it to you? Let’s look at f (x) = x2 . Recall (x) . This is exactly that in the first week of class, I had you calculate something that looked like f (x+h)−f h where we use that idea, except we want to find the instantaneous slope of a curve. In particular, we are going to take a limit as h gets really small. In proper notation, we are taking lim h→0 f (x + h) − f (x) h This is the definition of a derivative. So, with f (x) = x2 , f (x + h) = (x + h)2 = x2 + 2xh + h2 so x2 + 2xh + h2 − x2 2xh + h2 f (x + h) − f (x) = lim = lim = lim (2x + h) = 2x h→0 h→0 h→0 h→0 h h h lim This is our instantaneous slope. So, for example for exercise 4, we have x = 1 so the slope at (1, 1) is 2x = 2(1) = 2. Now, we just want to find the equation of a line with slope 2 that goes through the point (1, 1). Using the form y = mx + b, with m = 2, x = 1 and y = 1, we just need to find b. Plugging in the numbers yields b = −1 so that the equation of our tangent line is y = 2x − 1. Let’s try this for a couple of problems! 6. Find the equation of the tangent line to the graph y = x3 + 2x at (1, 3). 7. Find the equation of the tangent line to the graph y = 1 x at (1, 1). 8. (Harder) See if you can show that the instantaneous slope of the graph y = xn for any integer n is nxn−1 . Hint: see if you can find a pattern (you will use something we call induction). This is called the power rule. So, the derivative of a function f (x) = xn with respect to x is given by f 0 (x) = nxn−1 . Try to find the instantaneous slope of the following functions using (a) the definition of a derivative and (b) the power rule. Thankfully, now that we’ve proved (or accepted) that the rule works in full generality, it’ll become a lot easier to compute many derivatives. 9. f (x) = 3x + 7 10. g(x) = 3x2 + 6x Challenge Exercises: 1. Let a, b, c be positive real numbers such that abc = 1. Show that a2 + b2 + c2 ≤ a3 + b3 + c3 using derivatives. (Hint: you may use the fact that if f (x) = nx for n some number, f 0 (x) = nx ln n. Another hint is: consider f (x) = ax + bx + cx .) 2. Prove the product rule (f · g)0 = f 0 · g + f · g 0 by using the definition of a derivative.
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