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 am trying 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 sets of equations:
\(\displaystyle x_2 x_6^{k_1} + x_3 x_6^{k_2}-x_1 a &=& -b\\\)
\(\displaystyle k_1 x_2 x_6^{k_1} + k_2 x_3 x_6^{k_2}&=&0\\\)
\(\displaystyle x_2 x_7^{k_1} + x_3 x_7^{k_2}+ x_7 -x_1 a &=& -c\\\)
\(\displaystyle k_1 x_2 x_7^{k_1} + k_2 x_3 x_7^{k_2}&=& -1\\\)
\(\displaystyle x_4 x_6^{k_1} + x_5 x_6^{k_2}+a x_1 &=& b\\\)
\(\displaystyle x_4 x_7^{k_1} + x_5 x_7^{k_2}+ a x_1&=& x_7 -c\\\)
\(\displaystyle x_4 y^{k_1} + x_5 y^{k_2}+ a x_1-e &=&y-c\\\)
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 \(\displaystyle $ \frac{d x_2}{d y}, \frac{d x_3}{d y},\frac{d x_1}{d y}$\)
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 \(\displaystyle $ \frac{d x_2}{d y}, \frac{d x_3}{d y},\frac{d x_1}{d y}$\). I can solve for them and plug back into the expression I am trying to calculate, 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
I am trying 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 sets of equations:
\(\displaystyle x_2 x_6^{k_1} + x_3 x_6^{k_2}-x_1 a &=& -b\\\)
\(\displaystyle k_1 x_2 x_6^{k_1} + k_2 x_3 x_6^{k_2}&=&0\\\)
\(\displaystyle x_2 x_7^{k_1} + x_3 x_7^{k_2}+ x_7 -x_1 a &=& -c\\\)
\(\displaystyle k_1 x_2 x_7^{k_1} + k_2 x_3 x_7^{k_2}&=& -1\\\)
\(\displaystyle x_4 x_6^{k_1} + x_5 x_6^{k_2}+a x_1 &=& b\\\)
\(\displaystyle x_4 x_7^{k_1} + x_5 x_7^{k_2}+ a x_1&=& x_7 -c\\\)
\(\displaystyle x_4 y^{k_1} + x_5 y^{k_2}+ a x_1-e &=&y-c\\\)
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 \(\displaystyle $ \frac{d x_2}{d y}, \frac{d x_3}{d y},\frac{d x_1}{d y}$\)
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 \(\displaystyle $ \frac{d x_2}{d y}, \frac{d x_3}{d y},\frac{d x_1}{d y}$\). I can solve for them and plug back into the expression I am trying to calculate, 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