12. Functions and Derivatives Objectives: What is a derivative. Examples: derivatives from first principles. Review of rules for differentiation. Using the Rules. Refs: B&Z 8.1, 3, 4, 5.

What is a derivative? Let’s consider any old function y=f(x) and draw the secant line through two points on its graph. y x x+h f(x) f(x+h) secant line f(x+h)-f(x) We calculate the slope of the secant line = Rise/Run f(x+h)-f(x) h  h So this is tan .

Now choose h1< h. We calculate the slope of the y x x+h f(x) f(x+h) h1 f(x+h1)-f(x) 1 f(x+h1) x+h1 We calculate the slope of the secant line = Rise/Run f(x+h1)-f(x) h1 So this is tan 1.

Now choose h2< h1. We calculate the slope of the y x x+h f(x) f(x+h) h2 f(x+h2)-f(x) 2 f(x+h1) x+h1 f(x+h2) x+h2 We calculate the slope of the secant line = Rise/Run f(x+h2)-f(x) h2 So this is tan 2.

If y = f(x) we define the derivative of f at x, denoted Now we make the distance between the two points very small - that is, we let h approach zero. In this way, we can calculate the slope of the tangent line at the point (x, f(x)). y x f(x) f(x+h) tangent to curve If y = f(x) we define the derivative of f at x, denoted by f '(x) to be:

Examples 1: We also use other notations for f '(x): Finding derivatives from first principles. (i) Find the slope of y=x2 at x=1. Using the formula:

So f ’(1) =2 ; that is, the tangent to y=x2 at x=1 has slope 2.

(ii) Find the slope of the line tangent to y=x2 for any point x.

Example 2: Find the slope of y=√x (x > 0).

See handout for details. We seldom need to find derivatives from first principles. As long as we know the derivatives of standard functions and the rules for differentiation we can tackle most problems we are likely to encounter. Rules for differentiation: See handout for details.

Examples 1. f (x) = x4 + x2 + x + 2 f ’(x) = 4x3 + 2x + 1 + 0 f (x) = loge(8x) => f (x) = loge(8) + loge(x) so f ’(x) = 0 + 1/x = 1/x

Examples 3. f (x) = 2e3x f ’(x) = 2(3e3x) = 6e3x 4. f (x) = 2 cos(x) + sin (3x) f ’(x) = -2sin(x) + 3cos(3x)

Examples 3 (Harder) 1. f (x) = x3loge(x). Let u=x3 and v=loge(x). Then u ’ = 3x2 and v ’ =1/x. f ’(x) = u ’ v + u v ’ = 3x2 loge(x) + x3 .1/x = 3x2 loge(x) + x2 2. f (x) = (e3x-4)(x2 + 4x +7) Let u= e3x-4 and v= x2 + 4x +7. Then u ’ = 3 e3x and v ’ = 2x +4. f ’(x) = u ’ v + u v ’ = 3 e3x (x2 + 4x +7) + (e3x-4)(2x +4) = e3x(3 x2 + 12x + 21 +2x +4) -8x-16 = e3x(3 x2 + 14x + 25) -8(x+2)

3. Let u=sin(2x) and v=x3. Then u ’ = 2cos(2x) and v ’ =3 x2.

4. Let u=ex + 3cos(x) and v=sin(x). Then u ’= ex -3sin(x) and v ’ =cos(x).

You should now be able to attempt Q’s 1-4 from Example Sheet 5 from the Orange Book.