# 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).

Let $z\left(x,y\right)={e}^{3xy},x\left(p,q\right)=\frac{p}{q}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}y\left(p,q\right)=\frac{q}{p}$ are functions. Use multivariable chain rule of partial derivatives to find
(i) $\frac{\partial z}{\partial p}$
(ii) $\frac{\partial z}{\partial q}$.
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(i) $\frac{\partial z}{\partial p}=\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial p}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial p}$ (1)
$⇒\because \frac{\partial z}{\partial x}={e}^{3xy}\cdot 3y⇒\frac{\partial z}{\partial x}=3y{e}^{3xy}$
$\frac{\partial z}{\partial y}={e}^{3xy}\cdot 3x⇒\frac{\partial z}{\partial y}=3x{e}^{3xy}$
$\frac{\partial x}{\partial p}=\frac{1}{q}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial x}{\partial q}=-\frac{9}{{q}^{2}}$
$\frac{\partial y}{\partial p}=-\frac{1}{{p}^{2}}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial y}{\partial p}=\frac{1}{p}$
Using these values equation (1) becomes:
$\frac{\partial z}{\partial p}=3y{e}^{3xy}\cdot \frac{1}{q}+3x{e}^{3xy}\left(-\frac{q}{{p}^{2}}\right)$
$\frac{\partial z}{\partial p}=\frac{3y{e}^{3xy}}{q}-\left(3xq\frac{{e}^{3xy}}{{p}^{2}}\right)$
$\frac{\partial z}{\partial p}=\left(3\frac{{e}^{3xy}}{q{p}^{2}}\left({p}^{2}y-x{q}^{2}\right)\right)$
(ii) $\frac{\partial z}{\partial q}=\frac{\partial z}{\partial x}\cdot \frac{\partial x}{\partial q}+\frac{\partial z}{\partial y}\cdot \frac{\partial y}{\partial q}$
$\frac{\partial z}{\partial q}=3y{e}^{3xy}\left(-\frac{p}{{q}^{2}}\right)+3x{e}^{3xy}\left(\frac{1}{p}\right)$
$\frac{\partial z}{\partial q}=\left(-3yp\frac{{e}^{3yx}}{{q}^{2}}+\frac{3x}{p}{e}^{3xy}\right)$
$\frac{\partial z}{\partial q}=\left(3\frac{{e}^{3xy}}{p{q}^{2}}\left(x{q}^{2}-y{p}^{2}\right)\right)$