# Total Differentiation of implicitly defined function with many variables

• May 18th 2011, 03:50 PM
Total Differentiation of implicitly defined function with many variables
I need help sorting out some basic calculus. I need to take the total derivate of the following function with respect to the variable $\displaystyle$y$$. Note, the variables are y and \displaystyle x_i$$ where $\displaystyle$i$$goes from 1 to 7, while all other symbols are constants: I want to calculate the total derivative of \displaystyle x_2 y^{k_1} + x_3 y^{k_2}-x_1 with respect to the variable \displaystyle y Where \displaystyle  x_1, x_2, x_3$$ satisfy the following set of equation seven equations in eight variables$\displaystyle x_1,x_2,x_3,x_4,x_5,x_6,x_7$and$\displaystyle y$. Assuming these equations have a solution and it is unique, what is happening is that I get to choose $\displaystyle y$ and for every choice of $\displaystyle y$ we get some values for $\displaystyle x_1,x_2,x_3,x_4,x_5,x_6,x_7$. The total derivative I need to calculate is the first order condition I need to figure out an optimal choice of $\displaystyle y$. It is easy to verify in the context of the problem that my solution is not a boundary solution. So anyway, $\displaystyle$ x_1, x_2, x_3 satisfy the following set of equation seven equations in eight variables$\displaystyle x_1,x_2,x_3,x_4,x_5,x_6,x_7$and$\displaystyle y$. :

$\displaystyle x_2 x_6^{k_1} + x_3 x_6^{k_2}-x_1 a &=& -b\\$
$\displaystyle k_1 x_2 x_6^{k_1} + k_2 x_3 x_6^{k_2}&=&0\\$
$\displaystyle x_2 x_7^{k_1} + x_3 x_7^{k_2}+ x_7 -x_1 a &=& -c\\$
$\displaystyle k_1 x_2 x_7^{k_1} + k_2 x_3 x_7^{k_2}&=& -1\\$
$\displaystyle x_4 x_6^{k_1} + x_5 x_6^{k_2}+a x_1 &=& b\\$
$\displaystyle x_4 x_7^{k_1} + x_5 x_7^{k_2}+ a x_1&=& x_7 -c\\$
$\displaystyle x_4 y^{k_1} + x_5 y^{k_2}+ a x_1-e &=&y-c\\$

This is how I have solved this:

Need to compute:

$\displaystyle \frac{d x_2}{d y} y^{k_1}+ x_2 k_1 y^{k_1-1} + \frac{d x_3}{d y} y^{k_2}+ x_3 k_2 y^{k_2-1} -\frac{d x_1}{d y}$

I will take the total derivative of the constraints to solve for $\displaystyle \frac{d x_2}{d y}$, $\displaystyle \frac{d x_3}{d y}$,$\displaystyle \frac{d x_1}{d y}$
So, we get the following equations:

$\displaystyle \frac{d x_2}{dy}x_6^{k_1}+ x_2 k_1 x_6^{k_1-1} \frac{d x_6}{d y} + \frac{d x_3}{dy}x_6^{k_1}+ x_3 k_2 x_6^{k_2-1} \frac{d x_6}{d y}-a \frac{d x_1}{dy} &=&0\\$
$\displaystyle k_1 (\frac{d x_2}{dy}x_6^{k_1}+ x_2 k_1 x_6^{k_1-1} \frac{d x_6}{d y}) + k_2 (\frac{d x_3}{dy}x_6^{k_1}+ x_3 k_2 x_6^{k_2-1} \frac{d x_6}{d y})&=&0\\$
$\displaystyle \frac{dx_2}{dy}x_7^{k_1}+ x_2 k_1 x_7^{k_1-1} \frac{dx_7}{dy}+\frac{dx_3}{dy}x_7^{k_2}+ x_3 k_2 x_7^{k_2-1} \frac{dx_7}{dy} + \frac{dx_7}{dy} - a \frac{d x_1}{dy}&=&0 \\$
$\displaystyle k_1 (\frac{d x_2}{dy}x_7^{k_1}+ x_2 k_1 x_7^{k_1-1} \frac{d x_7}{d y}) + k_2 (\frac{d x_3}{dy}x_7^{k_1}+ x_3 k_2 x_7^{k_2-1} \frac{d x_7}{d y})&=&0\\$
$\displaystyle \frac{d x_4}{dy}x_6^{k_1}+ x_4 k_1 x_6^{k_1-1} \frac{d x_6}{d y} + \frac{d x_5}{dy}x_6^{k_1}+ x_5 k_2 x_6^{k_2-1} \frac{d x_6}{d y}+ a \frac{d x_1}{dy} &=& 0\\$
$\displaystyle \frac{d x_4}{dy}x_7^{k_1}+ x_4 k_1 x_7^{k_1-1} \frac{d x_7}{d y} + \frac{d x_5}{dy}x_7^{k_1}+ x_5 k_2 x_7^{k_2-1} \frac{d x_7}{d y}+ a \frac{d x_1}{dy} &=& \frac{d x_7}{dy}\\$
$\displaystyle \frac{d x_4}{d y} y^{k_1}+ x_4 k_1 y^{k_1-1} + \frac{d x_5}{d y} y^{k_2}+ x_5 k_2 y^{k_2-1}+ a \frac{d x_1}{dy}&=& 1\\$
These equations are linear in $\displaystyle \frac{d x_2}{d y}$, $\displaystyle \frac{d x_3}{d y}$,$\displaystyle \frac{d x_1}{d y}$. I can solve for them and plug back into ($\ref{eq:main}$) to get the final answer.

Is this correct? or am I doing something wrong? If this is correct, I'd appreciate it if you can recommend some reading I could read which covers implicit differentiation in more than two variables so I can understand how to think about this without having doubts. If this is incorrect, then could you please tell me why and then again recommend some place I can read about this. Thanks