# Let z(x,y)=e^(3xy), x(p,q)=p/q and y(p,q)=q/p are functions. Use multivariable chain rule of partial derivatives to find (i) (delz)/(delp) (ii) (delz)/(delq).

Question
Multivariable functions
Let $$\displaystyle{z}{\left({x},{y}\right)}={e}^{{{3}{x}{y}}},{x}{\left({p},{q}\right)}=\frac{{p}}{{q}}{\quad\text{and}\quad}{y}{\left({p},{q}\right)}=\frac{{q}}{{p}}$$ are functions. Use multivariable chain rule of partial derivatives to find
(i) $$\displaystyle\frac{{\partial{z}}}{{\partial{p}}}$$
(ii) $$\displaystyle\frac{{\partial{z}}}{{\partial{q}}}$$.

2021-01-07
(i) $$\displaystyle\frac{{\partial{z}}}{{\partial{p}}}=\frac{{\partial{z}}}{{\partial{x}}}\cdot\frac{{\partial{x}}}{{\partial{p}}}+\frac{{\partial{z}}}{{\partial{y}}}\cdot\frac{{\partial{y}}}{{\partial{p}}}$$ (1)
$$\displaystyle\Rightarrow\because\frac{{\partial{z}}}{{\partial{x}}}={e}^{{{3}{x}{y}}}\cdot{3}{y}\Rightarrow\frac{{\partial{z}}}{{\partial{x}}}={3}{y}{e}^{{{3}{x}{y}}}$$
$$\displaystyle\frac{{\partial{z}}}{{\partial{y}}}={e}^{{{3}{x}{y}}}\cdot{3}{x}\Rightarrow\frac{{\partial{z}}}{{\partial{y}}}={3}{x}{e}^{{{3}{x}{y}}}$$
$$\displaystyle\frac{{\partial{x}}}{{\partial{p}}}=\frac{{1}}{{q}}{\quad\text{and}\quad}\frac{{\partial{x}}}{{\partial{q}}}=-\frac{{9}}{{q}^{{2}}}$$
$$\displaystyle\frac{{\partial{y}}}{{\partial{p}}}=-\frac{{1}}{{p}^{{2}}}{\quad\text{and}\quad}\frac{{\partial{y}}}{{\partial{p}}}=\frac{{1}}{{p}}$$
Using these values equation (1) becomes:
$$\displaystyle\frac{{\partial{z}}}{{\partial{p}}}={3}{y}{e}^{{{3}{x}{y}}}\cdot\frac{{1}}{{q}}+{3}{x}{e}^{{{3}{x}{y}}}{\left(-\frac{{q}}{{p}^{{2}}}\right)}$$
$$\displaystyle\frac{{\partial{z}}}{{\partial{p}}}=\frac{{{3}{y}{e}^{{{3}{x}{y}}}}}{{q}}-{\left({3}{x}{q}\frac{{e}^{{{3}{x}{y}}}}{{p}^{{2}}}\right.}$$
$$\displaystyle\frac{{\partial{z}}}{{\partial{p}}}={\left({3}\frac{{e}^{{{3}{x}{y}}}}{{{q}{p}^{{2}}}}{\left({p}^{{2}}{y}-{x}{q}^{{2}}\right)}\right.}$$
(ii) $$\displaystyle\frac{{\partial{z}}}{{\partial{q}}}=\frac{{\partial{z}}}{{\partial{x}}}\cdot\frac{{\partial{x}}}{{\partial{q}}}+\frac{{\partial{z}}}{{\partial{y}}}\cdot\frac{{\partial{y}}}{{\partial{q}}}$$
$$\displaystyle\frac{{\partial{z}}}{{\partial{q}}}={3}{y}{e}^{{{3}{x}{y}}}{\left(-\frac{{p}}{{q}^{{2}}}\right)}+{3}{x}{e}^{{{3}{x}{y}}}{\left(\frac{{1}}{{p}}\right)}$$
$$\displaystyle\frac{{\partial{z}}}{{\partial{q}}}={\left(-{3}{y}{p}\frac{{e}^{{{3}{y}{x}}}}{{q}^{{2}}}+\frac{{{3}{x}}}{{p}}{e}^{{{3}{x}{y}}}\right)}$$
$$\displaystyle\frac{{\partial{z}}}{{\partial{q}}}={\left({3}\frac{{e}^{{{3}{x}{y}}}}{{{p}{q}^{{2}}}}{\left({x}{q}^{{2}}-{y}{p}^{{2}}\right)}\right.}$$

### Relevant Questions

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a) $$\displaystyle\frac{{\partial{k}}}{{\partial{a}}}$$
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