Multivariate functions: Partial derivatives
Elasticity in two variables
Recall that, for a positive differentiable function #f# in one variable, the elasticity for #x \gt 0# is defined as:
\[{\rm El}_xf(x)=f'(x) \cdot \dfrac{x}{f(x)}\tiny.\]
Here we treat the two variable case.
Elasticity
Let #f# be a bivariate function with positive values defined on a domain inside #\{\rv{x,y}\in\mathbb{R}^2\mid x\ge0\land y\ge0\}#. The elasticity of #f# with respect to #x# is defined as:
\[{\rm El}_{x}f(x,y)=f_x(x,y) \cdot \dfrac{x}{f(x,y)}\tiny.\]
The elasticity of #f# with respect to #y# is defined as \[{\rm El}_y{f}(x,y)=f_y(x,y) \cdot \dfrac{y}{f(x,y)}\tiny.\]
#{\rm El}_{x}z(x,y)=# #{{2\cdot x^4}\over{y^4+x^4}}#
#{\rm El}_yz(x,y)=# #{{2\cdot y^4}\over{y^4+x^4}}#
The elasticity #{\rm El}_xz(x,y)# with respect to #x# can be calculated as follows: first compute the partial derivative of #z# with respect to #x#:
\[ \begin{array}{rcl}
z_x (x,y)&=&\displaystyle\frac{\partial}{\partial x}\left( \sqrt{y^4+x^4} \right)\\
&&\phantom{xxx}\color{blue}{\text{function rule of }z}\\
&=&\displaystyle {{2\cdot x^3}\over{\sqrt{y^4+x^4}}} \\
&&\phantom{xxx}\color{blue}{\text{partial derivative calculated}}\\
\end{array}\]
Now
\[ \begin{array}{rcl}
{\rm El}_{x}z(x,y)& =&z_x(x,y) \cdot \dfrac{x}{z(x,y)}\\
&&\phantom{xxx}\color{blue}{\text{definition of elasticity}}\\
& =&\displaystyle\frac{x}{\sqrt{y^4+x^4}}\cdot \left({{2\cdot x^3}\over{\sqrt{y^4+x^4}}}\right)\\
&&\phantom{xxx}\color{blue}{\text{function rules substituted}}\\
&=& \displaystyle {{2\cdot x^4}\over{y^4+x^4}}\\
&&\phantom{xxx}\color{blue}{\text{simplified}}\\
\end{array}\]
The calculation of #{\rm El}_{y}z(x,y)# is similar: first compute the partial derivative of #z# with respect to #y#:
\[\begin{array}{rcl}
z_y(x,y) &=& \displaystyle {{2\cdot y^3}\over{\sqrt{y^4+x^4}}} \end{array}\]
Now
\[\begin{array}{rcl}
{\rm El}_{y}(x,y)& =&z_y(x,y) \cdot \dfrac{y}{z(x,y)}\\
&&\phantom{xxx}\color{blue}{\text{definition of elasticity}}\\
& =&\displaystyle\frac{y}{\sqrt{y^4+x^4}}\cdot\left( {{2\cdot y^3}\over{\sqrt{y^4+x^4}}}\right)\\
&&\phantom{xxx}\color{blue}{\text{function rules substituted}}\\
&=& \displaystyle {{2\cdot y^4}\over{y^4+x^4}}\\
&&\phantom{xxx}\color{blue}{\text{simplified}}\\
\end{array} \]
Or visit omptest.org if jou are taking an OMPT exam.