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

Activity6.3Not Simplifying First

As always, you want to simplify an expression before jumping in to take its derivative. Nevertheless, it can build confidence to see that the rules work even if you don't simplify first.

Subsection6.3.1Exercises

Consider the functions \(\fe{f}{x}=\sqrt{x^2}\) and \(\fe{g}{x}=\left(\sqrt{x}\right)^2\text{.}\)

1

Assuming that \(x\) is not negative, how does each of these formulas simplify? Use the simplified formula to find the formulas for \(\fe{\fd{f}}{x}\) and \(\fe{\fd{g}}{x}\text{.}\)

2

Use the chain rule (without first simplifying) to find the formulas for \(\fe{\fd{f}}{x}\) and \(\fe{\fd{g}}{x}\text{;}\) simplify each result (assuming that \(x\) is positive).

3

Are \(f\) and \(g\) the same function? Explain why or why not.

4

So long as \(x\) falls on the interval \(\ointerval{-\frac{\pi}{2}}{\frac{\pi}{2}}\text{,}\) \(\fe{\tan^{-1}}{\fe{\tan}{x}}=x\text{.}\) Use the chain rule to find \(\lzoo{x}{\fe{\tan^{-1}}{\fe{\tan}{x}}}\) and show that it simplifies as it should.

5

How does the formula \(\fe{g}{t}=\fe{\ln}{e^{5t}}\) simplify and what does this tell you about the formula for \(\fe{\fd{g}}{t}\text{?}\) After answering those questions, use the chain rule to find the formula for \(\fe{\fd{g}}{t}\) and show that it simplifies as it should.

Consider the function \(\fe{g}{t}=\fe{\ln}{\frac{5}{t^3\fe{\sec}{t}}}\text{.}\)

6

Find the formula for \(\fe{\fd{g}}{t}\) without first simplifying the formula for \(\fe{g}{t}\text{.}\)

7

Use the quotient, product, and power rules of logarithms to expand the formula for \(\fe{g}{t}\) into three logarithmic terms. Then find \(\fe{\fd{g}}{t}\) by taking the derivative of the expanded version of \(g\text{.}\)

8

Show the two resultant fomulas are in fact the same. Also, reflect upon which process of differentiation was less work and easier to “clean up.”