Thursday, March 19, 2009

Solved Problem on Partial Differentiation

Topic : Partial Differentiation
Problem : Solve the function by differentiation f(x,y) = xe^(x^2 y) at the intervals (1, ln 2)

Solution:
Following are the steps to differentiate a given function.

f(x,y) = xe^{x^2y}\\\frac{\partial f}{\partial x}=? And \frac{\partial f}{\partial y}=?\\\frac{\partial f}{\partial x}_{(1,2)}=? And \frac{\partial f}{\partial y}_{(1,ln 2)}=?\\ f(x,y) = xe^{x^2y}<br />\\Consider \frac{\partial e^{x^2y}}{\partial x}\\\frac{\partial e^u}{\partial x} and u=x^2y\\\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\ as u=x^2y\\\frac{\partial e^u}{\partial u}=e^u=e^{x^2y}----------(a)\\\frac{\partial u}{\partial x}=2xy-------(b)\\ By Chain Rule\\\frac{\partial e^{x^2y}}{\partial x}=\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\=e^{x^2y} . 2xy where a = e^{x^2y} and b=2xy\\\frac{\partial f}{\partial x}=x \frac{\partial e^{x^2y}}{\partial x}+e^{x^2y}\frac{\partial x}{\partial x}\\=xe^{x^2y}(2xy)+e^{x^2y}(1)\\=e^{x^2y}[2x^2y+1]\\\frac{\partial f}{\partial x}=xe^{x^2y}(x^2)\\=x^3e^{x^2y}\\Therefore\frac{\partial f}{\partial x}_{(1,ln2)}=e^{1ln2}(i^2.ln2)=e^{ln2}=2ln2\\\frac{\partial f}{\partial y}_{(1,ln2)}=1^3 e^{1^2ln(2)}=e^{ln(2)}=2<br />
at

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