Derivative-practice-chain-rule-Lagrange-notation.tex

\Section{Derivative practice---chain rule, Lagrange notation}

\begin{ProblemSet}[pencil space=2in]

 \begin{Problem}[pencil space=1in]
  What is $\OuterStyle{v'(u)}$?
  \begin{equation*}
   \OuterStyle{v(u) = \sqrt{u}}
  \end{equation*}
 \end{Problem}

 \begin{Problem}[pencil space=1in]
  What is $\InnerStyle{u'(x)}$?
  \begin{equation*}
   \InnerStyle{u(x) = 6 x^2 - 4x + 7}
  \end{equation*}
 \end{Problem}

 \begin{Problem}[pencil space=3in]
  What is $f'(x)$?
  \begin{equation*}
   f(x) = \OuterStyle{\sqrt{\InnerStyle{6 x^2 - 4x + 7}}}
  \end{equation*}
 \end{Problem}

  \begin{Problem}
  What is $g'(x)$?
  \begin{equation*}
   g(x) = \OuterStyle{\big(\InnerStyle{3 - 2x + 5x^2}\big)^2}
  \end{equation*}
 \end{Problem}

 \begin{Problem}
  What is $f'(x)$?
  \begin{equation*}
   f(x) = 4 \left(x^2 + 7\right)^5
  \end{equation*}
 \end{Problem}

 \begin{Problem}
  What is $f'(x)$?
  \begin{equation*}
   f(x) = \left(4 x^2 + 7\right)^5
  \end{equation*}
 \end{Problem}

 \begin{Problem}
  What is $f'(t)$?
  \begin{equation*}
   f(t) = \frac{1}{\sqrt{13 t}}
  \end{equation*}
 \end{Problem}

 \begin{Problem}
  What is $q'(t)$?
  \begin{equation*}
   q(t) = \left(t + \frac{3}{t}\right)^{\nicefrac{4}{3}}
  \end{equation*}
 \end{Problem}

 \begin{Problem}
  What is $h'(z)$?
  \begin{equation*}
   h(z) = 16 \left(z^4 - z^5\right)^{-5} - 7 z^{-3}
  \end{equation*}
 \end{Problem}

 % \begin{Problem}
 %  What is $p'(x)$?
 %  \begin{equation*}
 %   p(x) = 10 (3 x + 4)^{\nicefrac{2}{3}} + 8 \sqrt{5 - x^2} + \sqrt{7}
 %  \end{equation*}
 % \end{Problem}

\end{ProblemSet}

%%% Local Variables:
%%% mode: latex
%%% TeX-master: "Business-calculus-workbook"
%%% End: