Lesson: Derivative Basics - 2 Obj – The Derivative Function

Differentiation – To find the derivative of a function. Derivative – The slope of the tangent line of a curve. - Instantaneous rate of change - y’, f’(x), dy/dx, d/dx

Ex. 1: Find the derivative, with respect to x, of f(x) = x2 + 1and use it to find the equation of the tangent line at x = 2.

Differentiability: It is possible that the limit that defines a derivative of a function, f, may not exist at certain points in the domain of f. When this occurs, the derivative at that point is undefined. If the derivative does exist at a point, x0, then the function is differentiable at x0.

In general, a function is differentiable at a point if the graph has a unique tangent line at the point. In other words, to be differentiable means that it has a derivative.

Conditions when a function is not differentiable. Corner Points Points of Vertical Tangency Cusps Why? At corner points, the slopes of the secant lines have different values from the right & left sides. Points of vertical tangency have slopes of secant lines with undefined slopes. Cusps have non equal slopes of on opposing sides.

Ex. 2: Given the graph of f(x) below, find the value of:

Recall that when finding , that you are Finding the derivative of the function, which is actually just the slope of the tangent line. So when you are asked to find , all you need to do is find the slope of the line tangent to the curve at x = -2. Draw a tangent line on your graph Find the slope: rise/run

Solution: Tangent line at x = -2 has a slope of 0.

Solution: Tangent line at x = -1 has a rise of about 2 And a run of about 1, so f ’(-1) = 2/1 = 2

Solution: Tangent line at x = -1 has a rise of about 1 And a run of about -2, so f ’(2) = 1/-2

Ex. 3: Match each graph in # a - f with the graph of its derivative, from A - F.

Example 3 solutions a - D b - F c - B d - C e - A f - E

Theorem 3.3.1 The derivative of a constant is zero d/dx [c] = 0 Ex. If f(x) = 7, then If g(x) = -8, then

The basis for the Power Rule Or my personal favorite version: The basis for the Power Rule In words: The derivative of a term of a function is the product of its exponent and coefficient, along with the variable raised to one less power.

Examples:

EX. 4: Find each derivative b) c) d) e) f)

Solutions to EX. 4: b) c)

EX. 4: Find each derivative

Examples:

EX. 6: An object is dropped from the Empire State Building that is modeled by the equation h(t) = 1250 - 16t2, where h(t) is measured in feet and t is in sec. Find the velocity function of the object. Find the instantaneous velocity at t = 5. Find the time interval over which the function is valid. D. What is the velocity of the object as it hits the ground?

A. B.

C. To determine the time interval for which the function is valid, we must examine possible values of the domain that makes sense. (The time the object is dropped and the time it hits the ground). The initial time is easy (t = 0). To find the final time, we need to locate when the object hits the ground. (when h(t) = 0)

So the function h(t)= 1250 – 16t2 is valid for the time interval [0,8.84] D. One could assume that the instantaneous velocity of the object when it hits the ground is zero, but we know the dangers of assuming, don’t we? The velocity is not zero until after it hits the ground. We need to examine the value from our velocity function when the object hits the ground.

:EX. 7: Let f(x) = 2x6 + x-9 Find EX. 8: Find dy/dx if EX. 9: If Find

EX. 10: At what points does the graph of have a horizontal tangent line? What do we know about the slope of a horizontal line? How can we use that information to find what we are looking for?

After substituting those points into the original equation, we get: and Therefore horizontal tangents exist at the points (1, 2) and (-1, 6).

EX. 11: If Find:

Normal line – The line perpendicular to the tangent line, at the point of tangency. Normal Line Tangent Line

Normal Line Tangent Line Secant Line

Ex. 12: Write the equation of the tangent line and the normal line to the function f(x) = 2x2 + 3x +5, at the point (-1, 4)