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2.1 Derivatives and Rates of Change

2.1 Derivatives and Rates of Change. The tangent problem. The slope of a line is given by:. The slope of the tangent to f(x)=x 2 at (1,1) can be approximated by the slope of the secant through (4,16):. We could get a better approximation if we move the point closer to (1,1), i.e. (3,9):.

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2.1 Derivatives and Rates of Change

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  1. 2.1 Derivatives and Rates of Change

  2. The tangent problem The slope of a line is given by: The slope of the tangent to f(x)=x2 at (1,1) can be approximated by the slope of the secant through (4,16): We could get a better approximation if we move the point closer to (1,1), i.e. (3,9): Even better would be the point (2,4):

  3. The tangent problem The slope of a line is given by: If we got really close to (1,1), say (1.1,1.21), the approximation would get better still How far can we go?

  4. The tangent problem slope at The slope of the curve at the point is: slope Note: This is the slope of the tangent line to the curve at the point.

  5. The velocity problem B distance (miles) A time (hours) (The velocity at one moment in time.) Consider a graph of displacement (distance traveled) vs. time. Average velocity can be found by taking: The speedometer in your car does not measure average velocity, but instantaneous velocity.

  6. Derivatives Definition: The derivative of a function at a number a, denoted by f ′(a), is if this limit exists. Example: Find f ′(a) for f(x)=x2+3.

  7. Equation of the tangent line The tangent line to y=f(x) at (a,f(a)) is the line through (a,f(a)) whose slope is equal to f ′(a). Then the equation of the tangent line to the curve y=f(x) at the point (a,f(a)): Example: Find an equation of the tangent line to f(x)=x2+3 at (1,4). From previous slide: f ′(1)=21=2. Thus, the equation is y-f(1)=f ′(1)(x-1) y-4=2(x-1) or y=2x+2

  8. Average rate of change = Instantaneous rate of change = Rates of Change: These definitions are true for any function. ( x does not have to represent time. )

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