# Partial Derivatives

• May 27th 2012, 01:17 PM
Kvandesterren
Partial Derivatives
$\begin{array}\\\text{Given:}\\
z=sin(2u)cos(3v)\\
\text{find}\frac{\partial{z}}{\partial{u}}\\
\\
\\
\\
[sin(2u)(0)]+[cos(3v)(cos(2u)(2)]\\
\\
\text{however, i'm not sure why the answer doesn't simply regard the cos(3v) as a constant, making the answer:}\\
\\
[cos(3v)][cos(2u)(2)],\\
\\
\text{as in the problem:}\\

f(x,y,z)=x^{2}y^{5}+xz^{2}\;\;\;\text{where}\;\;f_ {x}=y^{5}2x+z^{2}(1).\\
\text{Here, }y^{5}\text{and }z^{2}\text{are treated as if they were constants.}\\
\\
\text{I suspect the chain rule has something to do with it, but i just don't know. Could someone help to clarify this?}

\end{array}
" alt="\begin{array}\\\text{Given:}\\
z=sin(2u)cos(3v)\\
\text{find}\frac{\partial{z}}{\partial{u}}\\
\\
\\
\\
[sin(2u)(0)]+[cos(3v)(cos(2u)(2)]\\
\\
\text{however, i'm not sure why the answer doesn't simply regard the cos(3v) as a constant, making the answer:}\\
\\
[cos(3v)][cos(2u)(2)],\\
\\
\text{as in the problem:}\\

f(x,y,z)=x^{2}y^{5}+xz^{2}\;\;\;\text{where}\;\;f_ {x}=y^{5}2x+z^{2}(1).\\
\text{Here, }y^{5}\text{and }z^{2}\text{are treated as if they were constants.}\\
\\
\text{I suspect the chain rule has something to do with it, but i just don't know. Could someone help to clarify this?}

\end{array}
" />
• May 27th 2012, 01:18 PM
Kvandesterren
Re: Partial Derivatives
Sorry everybody. I tried to post in latex, but obviously I messed up.