# [Math] Implicit differentiation with 3 variables and 2 simultaneous equations

calculusimplicit-differentiationmultivariable-calculus

Translated from Danish:

The "equation system" $e^x+y^2+x=2\,\, \land\,\, z^2+xy=4$ implicitly defines $x$ and $z$ as functions $x=x(y)$ and $z=z(y)$ of the variable $y$ around the point $(x_0,y_0,z_0)=(0,1,2)$.

Determine

$$\frac {dx}{dy}(1)\,\,, \frac {dz}{dy}(1)$$

Just by looking at the first of the two equation, I have come to the conclusion that $\frac {dx}{dy}(1)=-1$, but I seem to get different results everytime I try to find $\frac {dz}{dy}(1)$. I have tried just looking at the second equation and treating $x$ as a constant and I have also tried setting the two equations equal to eachother.

First notice the following, where prime denotes derivative with respect to $y$

\eqalign{ & {e^x} + {y^2} + x = 2 \cr & x'{e^x} + 2y + x' = 0 \cr & \left( {{e^x} + 1} \right)x' = - 2y \cr & x' = - {{2y} \over {{e^x} + 1}} \cr} \tag{1}

and hence when $x=0,y=1$, $x'$ will be

$$x'(1) = - {2 \over {1 + 1}} = - 1\tag{2}$$

Now, we repeat the same process with the second equation

\eqalign{ & {z^2} + xy = 4 \cr & 2zz' + x'y + x = 0 \cr & z' = - {{x'y + x} \over {2z}} \cr}\tag{3}

according to $(2)$ and $x=0,y=1,z=2$ we have

$$z'(1) = - {{\left( { - 1} \right)\left( 1 \right) + 0} \over {2\left( 2 \right)}} = {1 \over 4}\tag{4}$$