# Find ∂w/∂u?

## Let w = x^2 + y^2 + z^2, x = uv, y = u cos(v), z = u sin(v). Use the chain rule to find ∂w/∂u when (u, v) = (9, 0).

Feb 19, 2018

$w = {x}^{2} + {y}^{2} + {z}^{2}$

$w = {\left(u v\right)}^{2} + {\left(u \cos v\right)}^{2} + {\left(u \sin v\right)}^{2}$

$\textcolor{w h i t e}{d}$

$\implies \frac{\mathrm{dw}}{\mathrm{du}} = 2 u v \cdot \left(u ' v + u v '\right) + 2 \left(u \cos v\right) \cdot \left(u ' \cos v + u \cos v '\right) + 2 \left(u \sin v\right) \cdot \left(u ' \sin v + u \sin v '\right)$

$\textcolor{w h i t e}{d}$

$\implies \frac{\mathrm{dw}}{\mathrm{du}} = 2 u v \cdot \left(v + u \frac{\mathrm{dv}}{\mathrm{du}}\right) + 2 \left(u \cos v\right) \cdot \left(\cos v - u \sin v \frac{\mathrm{dv}}{\mathrm{du}}\right) + 2 \left(u \sin v\right) \cdot \left(\sin v + u \cos v \frac{\mathrm{dv}}{\mathrm{du}}\right)$

$\textcolor{w h i t e}{d}$

$\implies \frac{\mathrm{dw}}{\mathrm{du}} = 2 u {v}^{2} + 2 {u}^{2} \frac{\mathrm{dv}}{\mathrm{du}} + 2 u {\cos}^{2} v - 2 {u}^{2} \cos v \sin v \frac{\mathrm{dv}}{\mathrm{du}} + 2 u {\sin}^{2} v + 2 {u}^{2} \sin v \cos v \frac{\mathrm{dv}}{\mathrm{du}}$

$\textcolor{w h i t e}{d}$

$\implies \frac{\mathrm{dw}}{\mathrm{du}} = 2 u {v}^{2} + 2 {u}^{2} \frac{\mathrm{dv}}{\mathrm{du}} + 2 u \left({\cos}^{2} v + {\sin}^{2} v\right)$

$\textcolor{w h i t e}{d}$

$\implies \frac{\mathrm{dw}}{\mathrm{du}} = 2 u {v}^{2} + 2 {u}^{2} \frac{\mathrm{dv}}{\mathrm{du}} + 2 u$