08-lesson.tex

\section*{Derivative Review. Powers, Sums, Differences}

Basic Formulas
\begin{equation}
  \begin{aligned}
    \frac{d}{dx} \left(f(x)+g(x)\right) &= \frac{d}{dx} f(x)+\frac{d}{dx} g(x) \\
    \frac{d}{dx} \left(f(x)-g(x)\right) &= \frac{d}{dx} f(x)-\frac{d}{dx} g(x) \\
    \frac{d}{dx} \left(af(x)\right) &= a\frac{d}{dx} f(x) \\
    \frac{d}{dx} a &= 0 \\
    \frac{d}{dx} x^a &= ax^{a-1} \\
  \end{aligned}
\end{equation}
where $a$ is a constant number. Last equation is true {\it even} if $a$ is not an integer.
  
\begin{questions}
\question
Find $f'(z)$ if
$$f(z)=z^7+5z^4-8z+2031$$
\begin{solution}[1.5in]
  Use all of the above rules:
  $$f'(z)=7z^6+20z^3-8$$
\end{solution}
\question
If $\displaystyle{g(x) = \sqrt{x}+\frac{1}{x}}$ find $g'(x)$
\begin{solution}[1.5in]
  Use all of the above rules:
  \begin{equation}
    \begin{aligned}
    g(x) &= x^{\frac{1}{2}}+ x^{-1}\cr
    \noalign{so}
    g'(x)&= \frac{1}{2}x^{-\frac{1}{2}} - x^{-2}cr
    &= \frac{1}{2}\frac{1}{\sqrt{x}} - \frac{1}{x^2}\cr
    &= \frac{1}{2}\frac{\sqrt{x}}{x} - \frac{1}{x^2}\cr
    \end{aligned}
  \end{equation}
\end{solution}
\question
Find $\displaystyle{\frac{d}{dy} y\sqrt[3]{y}}$, simplify then take the derivative, don't use the product rule.
\begin{solution}[1.5in]
  $$y\sqrt[3]{y}=y\cdot y^{\frac{1}{3}}=y^{\frac{4}{3}}$$
    so
    $$\frac{d}{dy} y\sqrt[3]{y} = \frac{d}{dy} y^{\frac{4}{3}} = \frac{4}{3}y^{\frac{1}{3}} = \frac{4}{3}\sqrt[3]{y}$$
\end{solution}
\newpage
\question
If $h(x)=x^{\sqrt{5}}$, find $h'(x)$. Hint: don't make it overcomplicated it is as simple as it seems.
\begin{solution}[1.5in]
  $$h'(x)=\sqrt{5}x^{\sqrt{5}-1}$$
\end{solution}
\question
If $g(w) = w^5+3w^3+\sqrt{17}w+\pi$, find $g''(w)$
\begin{solution}[1.5in]
  $$g'(w)=5w^4+9w^2+\sqrt{17},$$ so
  $$g''(w)=20w^3+18w$$
\end{solution}
\question
If $h(q) = q^{.37}$ find $h'(q)$
\begin{solution}[1.5in]
  $h'(q)=.37q^{-.63}$
\end{solution}
\end{questions}