Find $\frac{\partial x}{\partial u}$, if $u=x+y^2$, $v=y+z^2$, $w=z+x^2$.

Question

I tried very hard but couldn't make a relation out them. I cross checked question too but in the book it's given as above. Hints will be appreciated too. Thanks.

Answer 1

$\text{We have}$:

$u=x+y^2\tag 1$ $v=y+z^2\tag 2$ $w=z+x^2\tag 3$

$\text{Therefore,}$

$\begin{align}x &=u-y^2[\text{ from }(1)]\\&=u-(v-z^2)^2[\text{ from }(2)]\\&=u-[v-(w-x^2)^2]^2[\text{ from }(3)]\\&=u-[v-(w^2-2wx^2+x^4)]^2\\&=u-[v-w^2+2wx^2-x^4)]^2\end{align}$

$\begin{align}\implies\dfrac{\partial x}{\partial u}&=1-2(v-w^2+2wx^2-x^4)(2w\cdot 2x-4x^3)\dfrac{\partial x}{\partial u}\\&=1-2(v-w^2+2wx^2-x^4)(4wx-4x^3)\dfrac{\partial x}{\partial u}\end{align}$

$\begin{align}\implies\dfrac{\partial x}{\partial u}&=\dfrac{1}{1+2(v-w^2+2wx^2-x^4)(4wx-4x^3)}\\&=\dfrac{1}{1+8x(v-w^2+2wx^2-x^4)(w-x^2)}\\&=\dfrac{1}{1+8x[v-w^2+wx^2+wx^2-x^4]z}[\text{ from }(3)]\\&=\dfrac{1}{1+8x[v-w^2+x^2(w-x^2)]z}\\&=\dfrac{1}{1+8xz[v-w^2+x^2z]}[\text{ from }(3)]\end{align}$