exercise:3131a9ed9e: Difference between revisions
From Stochiki
(Created page with "<div class="d-none"><math> \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow}...") |
No edit summary |
||
| Line 31: | Line 31: | ||
\newcommand{\conj}[1]{\overline{#1}} | \newcommand{\conj}[1]{\overline{#1}} | ||
\newcommand{\mathds}{\mathbb} | \newcommand{\mathds}{\mathbb} | ||
\renewcommand{\d}{d} | |||
</math></div> | </math></div> | ||
If <math>x = 16t^2 + 2t</math> and <math>y = \frac1x</math> and | If <math>x = 16t^2 + 2t</math> and <math>y = \frac1x</math> and | ||
<math>z = y^2 + 1</math>, compute <math>\d z</math> in terms of <math>t</math> and <math>\d t</math>. | <math>z = y^2 + 1</math>, compute <math>\d z</math> in terms of <math>t</math> and <math>\d t</math>. | ||
Latest revision as of 01:25, 23 November 2024
[math]
\newcommand{\ex}[1]{\item }
\newcommand{\sx}{\item}
\newcommand{\x}{\sx}
\newcommand{\sxlab}[1]{}
\newcommand{\xlab}{\sxlab}
\newcommand{\prov}[1] {\quad #1}
\newcommand{\provx}[1] {\quad \mbox{#1}}
\newcommand{\intext}[1]{\quad \mbox{#1} \quad}
\newcommand{\R}{\mathrm{\bf R}}
\newcommand{\Q}{\mathrm{\bf Q}}
\newcommand{\Z}{\mathrm{\bf Z}}
\newcommand{\C}{\mathrm{\bf C}}
\newcommand{\dt}{\textbf}
\newcommand{\goesto}{\rightarrow}
\newcommand{\ddxof}[1]{\frac{d #1}{d x}}
\newcommand{\ddx}{\frac{d}{dx}}
\newcommand{\ddt}{\frac{d}{dt}}
\newcommand{\dydx}{\ddxof y}
\newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}}
\newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}}
\newcommand{\dist}{\mathrm{distance}}
\newcommand{\arccot}{\mathrm{arccot\:}}
\newcommand{\arccsc}{\mathrm{arccsc\:}}
\newcommand{\arcsec}{\mathrm{arcsec\:}}
\newcommand{\arctanh}{\mathrm{arctanh\:}}
\newcommand{\arcsinh}{\mathrm{arcsinh\:}}
\newcommand{\arccosh}{\mathrm{arccosh\:}}
\newcommand{\sech}{\mathrm{sech\:}}
\newcommand{\csch}{\mathrm{csch\:}}
\newcommand{\conj}[1]{\overline{#1}}
\newcommand{\mathds}{\mathbb}
\renewcommand{\d}{d}
[/math]
If [math]x = 16t^2 + 2t[/math] and [math]y = \frac1x[/math] and [math]z = y^2 + 1[/math], compute [math]\d z[/math] in terms of [math]t[/math] and [math]\d t[/math].