91Ó°ÊÓ

$\begin{array}{l}x=r\sin \mathrm{θ}\cos \mathrm{ϕ}\\ y=r\sin \mathrm{θ}\sin \mathrm{ϕ}\\ z=r\cos \mathrm{θ}\end{array}$

Also,

$r^2=x^2+y^2+z^2$and$\mathrm{θ}={\cos }^{−1}\left(\frac{z}{r}\right)$

The objective is to determine $\frac{∂\mathrm{θ}}{∂x}$and to show that$\frac{∂x}{∂\mathrm{θ}}≠\frac{∂\mathrm{θ}}{∂x}$ differentiate$x=r\sin \mathrm{θ}\cos \mathrm{ϕ}$with respect to $\mathrm{θ}$.

$\frac{∂x}{∂\mathrm{θ}}=r\cos \mathrm{θ}\cos \mathrm{ϕ}$take r and $\mathrm{ϕ}$as constants.

Thus, the value of $\frac{∂x}{∂\mathrm{θ}}$is$\frac{∂x}{∂\mathrm{θ}}=r\cos \mathrm{θ}\cos \mathrm{ϕ}\mathrm{...}\left(1\right)$

Differentiate$\mathrm{θ}={\cos }^{−1}\left(\frac{z}{r}\right)$with respect to.

$\begin{array}{c}\frac{∂\mathrm{θ}}{∂x}=\frac{∂\mathrm{θ}}{∂r}\frac{∂r}{∂x}\\ =\frac{z}{r^2\sqrt{1−\frac{z^2}{r^2}}}\frac{x}{\sqrt{x^2+y^2+z^2}}\\ =\frac{z}{r\sqrt{r^2−z^2}}\frac{x}{r}⇒(r=\sqrt{x^2+y^2+z^2})\\ =\frac{zx}{r^2\sqrt{r^2−z^2}}\end{array}$

Substitute x and z in the equation,

$\begin{array}{c}\frac{∂\mathrm{θ}}{∂x}=\frac{zx}{r^2\sqrt{r^2−z^2}}\\ \frac{∂\mathrm{θ}}{∂x}=\frac{\left(r\cos \mathrm{θ}\right)\left(r\sin \mathrm{θ}\cos \mathrm{ϕ}\right)}{r^2\sqrt{r^2−{\left(r\cos \mathrm{θ}\right)}^2}}\\ \frac{∂\mathrm{θ}}{∂x}=\frac{r^2\cos \mathrm{θ}\sin \mathrm{θ}\cos \mathrm{ϕ}}{r^2\sqrt{r^2−r^2{\cos }^2\mathrm{θ}}}\end{array}$

So, we get,

$\begin{array}{c}\frac{∂\mathrm{θ}}{∂x}=\frac{\cos \mathrm{θ}\sin \mathrm{θ}\cos \mathrm{ϕ}}{\sqrt{r^2(1−{\cos }^2\mathrm{θ})}}\\ \frac{∂\mathrm{θ}}{∂x}=\frac{\cos \mathrm{θ}\sin \mathrm{θ}\cos \mathrm{ϕ}}{\sqrt{r^2{\sin }^2\mathrm{θ}}}⇒({\cos }^2\mathrm{θ}+{\sin }^2\mathrm{θ}=1)\\ =\frac{\cos \mathrm{θ}\sin \mathrm{θ}\cos \mathrm{ϕ}}{r\sin \mathrm{θ}}\\ =\frac{\cos \mathrm{θ}\cos \mathrm{ϕ}}{r}\end{array}$

Thus, the value of $\frac{∂\mathrm{θ}}{∂x}$is$\frac{∂\mathrm{θ}}{∂x}=\frac{1}{r}\cos \mathrm{θ}\cos \mathrm{ϕ}\mathrm{...}\left(2\right)$

From equation (1) and (2) observe that,

$\frac{∂x}{∂\mathrm{θ}}≠\frac{∂\mathrm{θ}}{∂x}$