Explanation Given information: The given function is g ( t ) = 3 t 2 ( 2 t 2 − 1 ) 2 . Formula used: The formula to calculate the differential of a function h ( x ) = [ g ( x ) ] n by General Power Rule is, h ′ ( x ) = n [ g ( x ) ] n − 1 ⋅ g ′ ( x ) The formula to calculate the differential of a function h ( x ) = [ g ( x ) f ( x ) k ( x ) ] by Product Rule is, h ′ ( x ) = f ( x ) k ( x ) g ′ ( x ) + f ′ ( x ) g ( x ) k ( x ) + f ( x ) g ( x ) k ′ ( x ) Calculation: The given function is, g ( t ) = 3 t 2 ( 2 t 2 − 1 ) 2 (1) Differentiate equation (1) with respect to t on both sides to evaluate g ′ ( t ) . g ′ ( t ) = g ( t ) = ( 2 t 2 − 1 ) 2 d d t ( 3 t 2 ) + 3 t 2 d d t ( 2 t 2 − 1 ) 2 [ Use Product Rule ] = 6 t ( 2 t 2 − 1 ) 2 + 3 t 2 ( 4 t ) 2 ( 2 t 2 − 1 ) [ Use General Power Rule d d x ( x n ) = n x n − 1 , n > 0 ] = 6 t ( 2 t 2 − 1 ) [ 2 t 2 − 1 + 4 t 2 ] = 6 t ( 2 t 2 − 1 ) [ 6 t 2 − 1 ] Again differentiate above equation with respect to t on the both sides, to evaluate g ″ ( t )

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ISBN: 9781337825597

Author: Tan

Publisher: INTER CENG

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Differential Equations

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Chapter 3.5, Problem 9E

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To find: The first and second derivative of given function.

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Chapter 3.5, Problem 8E

Chapter 3.5, Problem 10E

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Consider the function f(x) = x²-1. (a) Find the instantaneous rate of change of f(x) at x=1 using the definition of the derivative. Show all your steps clearly. (b) Sketch the graph of f(x) around x = 1. Draw the secant line passing through the points on the graph where x 1 and x-> 1+h (for a small positive value of h, illustrate conceptually). Then, draw the tangent line to the graph at x=1. Explain how the slope of the tangent line relates to the value you found in part (a). (c) In a few sentences, explain what the instantaneous rate of change of f(x) at x = 1 represents in the context of the graph of f(x). How does the rate of change of this function vary at different points?

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The first and second derivative of given function.

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