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\(\newcommand{\Z}{\mathbb{Z}} \newcommand{\reals}{\mathbb{R}} \newcommand{\real}[1]{\mathbb{R}^{#1}} \newcommand{\fe}[2]{#1\mathopen{}\left(#2\right)\mathclose{}} \newcommand{\cinterval}[2]{\left[#1,#2\right]} \newcommand{\ointerval}[2]{\left(#1,#2\right)} \newcommand{\cointerval}[2]{\left[\left.#1,#2\right)\right.} \newcommand{\ocinterval}[2]{\left(\left.#1,#2\right]\right.} \newcommand{\point}[2]{\left(#1,#2\right)} \newcommand{\fd}[1]{#1'} \newcommand{\sd}[1]{#1''} \newcommand{\td}[1]{#1'''} \newcommand{\lz}[2]{\frac{d#1}{d#2}} \newcommand{\lzn}[3]{\frac{d^{#1}#2}{d#3^{#1}}} \newcommand{\lzo}[1]{\frac{d}{d#1}} \newcommand{\lzoo}[2]{{\frac{d}{d#1}}{\left(#2\right)}} \newcommand{\lzon}[2]{\frac{d^{#1}}{d#2^{#1}}} \newcommand{\lzoa}[3]{\left.{\frac{d#1}{d#2}}\right|_{#3}} \newcommand{\abs}[1]{\left|#1\right|} \newcommand{\sech}{\operatorname{sech}} \newcommand{\csch}{\operatorname{csch}} \newcommand{\lt}{ < } \newcommand{\gt}{ > } \newcommand{\amp}{ & } \)

Activity5.9Simplification

In Activity 5.7 Exercises 5.7.1.5–5.7.1.7 and Activity 5.8 Exercises 5.8.1.5–5.8.1.7 you applied the product and quotient rules to expressions where the derivative could have been found much more quickly had you simplified the expression before taking the derivative. For example, while you can find the correct derivative formula using the quotient rule when working Exercise 5.8.1.7, the derivative can be found much more quickly if you simplify the expression before applying the rules of differentiation. This is illustrated in Example 5.9.1.

Example5.9.1

\begin{align*} \lzoo{x}{\frac{10}{2x}}&=\lzoo{x}{5x^{-1}}\\ &=-5x^{-2}\\ &=-\frac{5}{x^{2}} \end{align*}

Part of learning to take derivatives is learning to make good choices about the methodology to employ when taking derivatives. In Examples 5.9.2 and 5.9.3, the need to use the product rule or quotient rule is obviated by first simplifying the expression being differentiated.

Example5.9.2

\begin{align*} \text{Find }\lz{y}{x}\text{ if }y&=\fe{\sec}{x}\fe{\cos}{x}\text{.}&y&=\fe{\sec}{x}\fe{\cos}{x}\\ &&&=\frac{1}{\fe{\cos}{x}}\fe{\cos}{x}\\ &&&=1\\ &&\lz{y}{x}&=0 \end{align*} Note that since \(y\) is undefined for \(x=\frac{\pi}{2}+k\pi\) with \(k\in\mathbb{Z}\), \(\lz{y}{x}\) is also undefined at such \(x\)-values.

Example5.9.3

\begin{align*} \text{Find }\fe{\fd{f}}{t}\text{ if }\fe{f}{t}&=\frac{4t^5-3t^3}{2t^2}\text{.}&\fe{f}{t}&=\frac{4t^5-3t^3}{2t^2}\\ &&&=\frac{4t^5}{2t^2}-\frac{3t^3}{2t^2}\\ &&&=2t^3-\frac{3}{2}t\\ &&\fe{\fd{f}}{t}&=6t^2-\frac{3}{2}\\ &&&=\frac{12t^2-3}{2} \end{align*} Note that since \(f\) is undefined at \(0\), \(\fd{f}\) is also undefined at \(0\).

Subsection5.9.1Exercises

Find the derivative with respect to \(x\) for each of the following functions after first completely simplifying the formula being differentiated. In each case you should not use either the product rule or the quotient rule while finding the derivative formula.

1

\(y=\dfrac{4x^{12}-5x^4+3x^2}{x^4}\)

2

\(\fe{g}{x}=\dfrac{-4\fe{\sin}{x}}{\fe{\cos}{x}}\)

3

\(\fe{h}{x}=\dfrac{4-x^6}{3x^{-2}}\)

4

\(\fe{z}{x}=\fe{\sin^2}{x}+\fe{\cos^2}{x}\)

5

\(z=(x+4)(x-4)\)

6

\(\fe{T}{x}=\dfrac{\fe{\ln}{x}}{\fe{\ln}{x^2}}\)