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Find values of a and b that make f ( x ) = { a x + b if x < 3 x 2 if x 3 both continuous and differentiable at 3 .

a = 6 and b = −9

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Higher-order derivatives

The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative. Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively, these are referred to as higher-order derivatives . The notation for the higher-order derivatives of y = f ( x ) can be expressed in any of the following forms:

f ( x ) , f ( x ) , f ( 4 ) ( x ) ,… , f ( n ) ( x )
y ( x ) , y ( x ) , y ( 4 ) ( x ) ,… , y ( n ) ( x )
d 2 y d x 2 , d 3 y d y 3 , d 4 y d y 4 ,… , d n y d y n .

It is interesting to note that the notation for d 2 y d x 2 may be viewed as an attempt to express d d x ( d y d x ) more compactly. Analogously, d d x ( d d x ( d y d x ) ) = d d x ( d 2 y d x 2 ) = d 3 y d x 3 .

Finding a second derivative

For f ( x ) = 2 x 2 3 x + 1 , find f ( x ) .

First find f ( x ) .

f ( x ) = lim h 0 ( 2 ( x + h ) 2 3 ( x + h ) + 1 ) ( 2 x 2 3 x + 1 ) h Substitute f ( x ) = 2 x 2 3 x + 1 and f ( x + h ) = 2 ( x + h ) 2 3 ( x + h ) + 1 into f ( x ) = lim h 0 f ( x + h ) f ( x ) h . = lim h 0 4 x h + h 2 3 h h Simplify the numerator. = lim h 0 ( 4 x + h 3 ) Factor out the h in the numerator and cancel with the h in the denominator. = 4 x 3 Take the limit.

Next, find f ( x ) by taking the derivative of f ( x ) = 4 x 3 .

f ( x ) = lim h 0 f ( x + h ) f ( x ) h Use f ( x ) = lim h 0 f ( x + h ) f ( x ) h with f ( x ) in place of f ( x ) . = lim h 0 ( 4 ( x + h ) 3 ) ( 4 x 3 ) h Substitute f ( x + h ) = 4 ( x + h ) 3 and f ( x ) = 4 x 3. = lim h 0 4 Simplify. = 4 Take the limit.
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Find f ( x ) for f ( x ) = x 2 .

f ( x ) = 2

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Finding acceleration

The position of a particle along a coordinate axis at time t (in seconds) is given by s ( t ) = 3 t 2 4 t + 1 (in meters). Find the function that describes its acceleration at time t .

Since v ( t ) = s ( t ) and a ( t ) = v ( t ) = s ( t ) , we begin by finding the derivative of s ( t ) :

s ( t ) = lim h 0 s ( t + h ) s ( t ) h = lim h 0 3 ( t + h ) 2 4 ( t + h ) + 1 ( 3 t 2 4 t + 1 ) h = 6 t 4.

Next,

s ( t ) = lim h 0 s ( t + h ) s ( t ) h = lim h 0 6 ( t + h ) 4 ( 6 t 4 ) h = 6.

Thus, a = 6 m/s 2 .

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For s ( t ) = t 3 , find a ( t ) .

a ( t ) = 6 t

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Key concepts

  • The derivative of a function f ( x ) is the function whose value at x is f ( x ) .
  • The graph of a derivative of a function f ( x ) is related to the graph of f ( x ) . Where f ( x ) has a tangent line with positive slope, f ( x ) > 0 . Where f ( x ) has a tangent line with negative slope, f ( x ) < 0 . Where f ( x ) has a horizontal tangent line, f ( x ) = 0 .
  • If a function is differentiable at a point, then it is continuous at that point. A function is not differentiable at a point if it is not continuous at the point, if it has a vertical tangent line at the point, or if the graph has a sharp corner or cusp.
  • Higher-order derivatives are derivatives of derivatives, from the second derivative to the n th derivative.

Key equations

  • The derivative function
    f ( x ) = lim h 0 f ( x + h ) f ( x ) h

For the following exercises, use the definition of a derivative to find f ( x ) .

f ( x ) = 2 3 x

−3

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f ( x ) = 1 x

−1 2 x 3 / 2

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For the following exercises, use the graph of y = f ( x ) to sketch the graph of its derivative f ( x ) .

Practice Key Terms 5

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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