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Exercise 34

Avg difficulty 2.5

In this table several function and derivative values are given for the functions ff and gg. The exercises that follow are based upon these two functions.

Various values for derivatives of ff and gg
xxf(x)\fe{f}{x}f′(x)\fe{\fd{f}}{x}f′′(x)\fe{\sd{f}}{x}g(x)\fe{g}{x}g′(x)\fe{\fd{g}}{x}g′′(x)\fe{\sd{g}}{x}
0022−1-1−2-2−3-3−3-3−2-2
11110044−5-5221616
2244771010111137375858
331515202016168787126126124124
44464639392222289289293293214214

Suppose that you were asked to find the value of h′(2)\fe{\fd{h}}{2} where h(x)=f(x)g(x)\fe{h}{x}=\fe{f}{x}\fe{g}{x}. The first thing you would need to do is find a formula for h′(x)\fe{\fd{h}}{x}. You would then replace xx with 22, substitute the appropriate function and derivative values, and simplify. This is illustrated below.h′(x)=ddx(f(x))g(x)+f(x)ddx(g(x))=f′(x)g(x)+f(x)g′(x)h′(2)=f′(2)g(2)+f(2)g′(2)=7(11)+4(37)=225\begin{aligned} \fe{\fd{h}}{x}&=\lzoo{x}{\fe{f}{x}}\fe{g}{x}+\fe{f}{x}\lzoo{x}{\fe{g}{x}} \\ &=\fe{\fd{f}}{x}\fe{g}{x}+\fe{f}{x}\fe{\fd{g}}{x} \\ \fe{\fd{h}}{2}&=\fe{\fd{f}}{2}\fe{g}{2}+\fe{f}{2}\fe{\fd{g}}{2} \\ &=7(11)+4(37) \\ &=225 \end{aligned}Following the process outlined in this example, find each of the following.

Find h′′(2)\fe{\sd{h}}{2}, where h(x)=f(x)g(x)\fe{h}{x}=\fe{f}{x}\fe{g}{x}.

Simonds, Calculus Lab Manual, §5.12 Supplement, Exercise 34

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Adapted from Calculus Lab Manual © 2012–2016 Portland Community College. Licensed under CC BY-SA 4.0. Source: Calculus Lab Manual. Imported at revision 3060f8e.