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Lecture 5 - Logarithms, Slope of a Function, Derivatives

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Lecture 5 - Logarithms, Slope of a Function, Derivatives Lecture 5 - Logarithms, Slope of a Function, Derivatives 5.1 Logarithms Note the graph of ex This graph passes the horizontal line test, so f(x) = ex is one-to-one and therefore has an inverse function. This is also true of f(x) = ax for any a > 0; a 6= 1. More generally, for any a > 1 the graph of ax and its inverse look like this. If f(x) = ax, then we define the inverse function f −1 to be the logarithm with base a, and write −1 f (x) = loga(x) x Note that, since the image of a is only the positive numbers, the domain of loga(x) is all positive real numbers. The key property is: b loga x = b () a = x Examples: ? log10 10 = 1 10 = 10 ? log5 25 = 2 5 = 25 1 1 ? 1 log4 2 = − 2 4 = 2 1 ? 1 log5 125 = −3 5 = 125 "" log equation corresponding exponential equation Log Rules 1. Most important: by the properties of inverse functions we have x logb x logb(b ) = x and b = x The most important case of logs is when b = e. Log base e has a special name, in fact we define loge x = ln(x). So the above becomes ln(ex) = x and eln(x) = x LEARN THIS!! The function ln(x) is known as the natural logarithm function, and ln(x) should be read as \the natural logarithm of x". In class, you may also hear me read this as \lawn x", but this isn't as standard. Other rules: (I will state for ln, but they work for every log). Suppose that x; y > 0 2. ln(xy) = ln(x) + ln(y) x 3. ln y = ln x − ln y 4. ln(xy) = y ln(x) Calculations: e3 ln(x) = eln(x3) = x3 1 ln = ln(e−1) = −1 e Rewrite the following: xy ln = ln(xy) − ln(z) = ln(x) + ln(y) − ln(z) z Compound Interest Revisited: (p324 #77) A deposit of $1000 is made in an account that earns interest at a rate of 5% per year. How long will it take for the balance to double if interest is compounded annually? solution: From our earlier formula, our balance after t years is :05t A(t) = 1000 1 + = 1000(1:05)t 1 We are trying to find t such that A(t) = 2000. So we set the formula equal to 2000: 2000 = 1000(1:05)t 2 = 1:05t We have to solve this for t. The general principle we use is that if we are trying to solve for a variable in the exponent, take log of both sides. So we get ln(2) = ln(1:05t) ln(2) = t ln(1:05) In this last point we see the point of using logs - the exponent can be brought down and solved for. So, ln(2) t = ≈ 14.21 years. ln(1:05) ln(2) Note that this is a very typical test/exam question. The answer I would expect is t = ln(1:05) . Example 2: Same question, but now interest is compounded 10 times a year. solution: By our formula, :0510t A(t) = 1000 1 + = 1000(1:005)10t 10 Again, we solve A(t) = 2000. 2000 = 1000(1:005)10t 2 = 1:00510t ln(2) = ln(1:00510t) ln(2) = 10t ln(1:005) Therefore, ln(2) t = ≈ 13:9 years. 10 ln(1:005) Example 3: Same question, but now interest is compounded continuously. solution: By our formula, A(t) = 1000e:05t Again, we solve A(t) = 2000. 2000 = 1000e:05t 2 = e:05t ln(2) = ln(e:05t) ln(2) = :05t Therefore, ln(2) t = ≈ 13:86 years. :05 Notice that continuous compounding gives a kind of \best case scenario" { no amount of compounding will get your money to double faster than approximately 13.86 years. 5.2 The Derivative and the Slope of Tangent Lines When given a graph of a function, we want to be able to calculate the slope of tangent lines. The slope of the tangent line at a point will tell us how rapidly the function is increasing or decreasing. We use a limit to find the slopes of tangent lines. A secant line for a function is one that intersects it at at least two points. The idea is to approximate the tangent line using secant lines. The slope of the tangent line will be the limit of the slopes of the secants, if it exists. The slope of the secant line connecting x1 to x is f(x1) − f(x) x1 − x If we pick a point near to x and denote it x + ∆x, the slope of the secant is f(x + ∆x) − f(x) f(x + ∆x) − f(x) = (x + ∆x) − x ∆x We should think of ∆x as being small, and note it can be positive or negative. So ideally the slope of the tangent line is f(x + ∆x) − f(x) lim ∆x!0 ∆x So we make the following definition. Definition 5.1 Let f(x) be a function. Then we define the derivative of f at x, denoted f 0(x), as f(x + ∆x) − f(x) f 0(x) = lim ∆x!0 ∆x if that limit exists, and if it does we say that f is differentiable at x. Notation: We will denote this a few different ways. If y = f(x), then all of the following denote the derivative: df d dy f 0(x); ; (f(x)); y0; : dx dx dx Examples: 1 1. Find the derivative of f(x) = 2 x + 3 at x = 2. solution: We calculate: f(2 + ∆x) − f(2) f 0(2) = lim ∆x!0 ∆x 1 (2 + ∆x) + 3 − ( 1 2 + 3) = lim 2 2 ∆x!0 ∆x 1 ∆x 1 = 2 = ∆x 2 1 1 So the slope of the tangent line is m = 2 . This makes sense because y = 2 x + 3 is a line, and the tangent line is the line itself. 2. Find the slope fo the tangent line for g(x) = 1 − x2 at x (unspecified). solution: We have g(x + ∆x) − g(x) g0(x) = lim ∆x!0 ∆x 1 − (x + ∆x)2 − (1 − x2) = lim ∆x!0 ∆x 1 − (x2 + 2x∆x + ∆x2) − (1 − x2) = lim ∆x!0 ∆x −2x∆x − ∆x2 = lim ∆x!0 ∆x = lim (−2x − ∆x) ∆x!0 = −2x So the slope of the tangent line at x is −2x. 1 3. Find the equation of the tangent line to f(x) = x at the point (1; 1). solution: First we have to find the slope, then we need to use it and the point (1; 1) to find the line's equation. f(x + ∆x) − f(x) f 0(x) = lim ∆x!0 ∆x 1 − 1 = lim x+∆x x ∆x!0 ∆x x − x+∆x = lim x(x+∆x) x(x+∆x) ∆x!0 ∆x x−(x+∆x) = lim x(x+∆x) ∆x!0 ∆x 1 (−∆x) = lim · ∆x!0 ∆x x(x + ∆x) −1 = lim ∆x!0 x2 + x∆x 1 = − x2 So f 0(1) = −1. Recall that y = mx + b, so here y = −x + b. We sub in the point (1; 1) to find b: 1 = −1 + 2 =) b = 2 =) y = −x + 2: Notes: 1. As you can see, calculating derivatives from the definition is difficult sometimes. We will develop rules to avoid this. 2. Not every function is differentiable everywhere. In fact, you can't take the derivative at points • of discontinuity, • with corners, • or with vertical tangents. Note that saying that a function isn't differentiable at points of discontinuity is equiv- alent to saying that differentiable at x =) continuous at x..
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