# How do you differentiate z=w^(3/2)(w+ce^w)?

Apr 28, 2017

$\textcolor{red}{\frac{\mathrm{dz}}{\mathrm{dw}} = \frac{\sqrt{w}}{2} \left[5 w + c {e}^{w} \left(2 w + 3\right)\right]}$

#### Explanation:

$z = {w}^{\frac{3}{2}} \left(w + c {e}^{w}\right)$

Assuming that c is a constant,

$\frac{\mathrm{dz}}{\mathrm{dw}} = \frac{d \left[{w}^{\frac{3}{2}} \left(w + c {e}^{w}\right)\right]}{\mathrm{dw}}$

Using chain rule

$\implies \frac{\mathrm{dz}}{\mathrm{dw}} = {w}^{\frac{3}{2}} \cdot \frac{d \left(w + c {e}^{w}\right)}{\mathrm{dw}} + \left(w + c {e}^{w}\right) \frac{d \left({w}^{\frac{3}{2}}\right)}{\mathrm{dw}}$

Using sum rule to evaluate $\frac{d \left(w + c {e}^{w}\right)}{\mathrm{dw}}$,

$\implies \frac{\mathrm{dz}}{\mathrm{dw}} = {w}^{\frac{3}{2}} \cdot \left[\frac{\mathrm{dw}}{\mathrm{dw}} + c \frac{{\mathrm{de}}^{w}}{\mathrm{dw}}\right] + \left(w + c {e}^{w}\right) \left(\frac{3}{2} {w}^{\frac{3}{2} - 1}\right)$

$\implies \frac{\mathrm{dz}}{\mathrm{dw}} = {w}^{\frac{3}{2}} \left(1 + c {e}^{w}\right) + \frac{3}{2} \left(w + c {e}^{w}\right) {w}^{\frac{1}{2}}$

$\implies \frac{\mathrm{dz}}{\mathrm{dw}} = {w}^{\frac{1}{2}} \left[w \left(1 + c {e}^{w}\right) + \frac{3}{2} w + \frac{3}{2} c {e}^{w}\right]$

$\implies \frac{\mathrm{dz}}{\mathrm{dw}} = \sqrt{w} \left[w + c \cdot w {e}^{w} + \frac{3}{2} w + \frac{3}{2} c {e}^{w}\right]$

$\implies \frac{\mathrm{dz}}{\mathrm{dw}} = \sqrt{w} \left[\frac{5}{2} w + \frac{1}{2} c {e}^{w} \left(2 w + 3\right)\right]$

$\implies \textcolor{red}{\frac{\mathrm{dz}}{\mathrm{dw}} = \frac{\sqrt{w}}{2} \left[5 w + c {e}^{w} \left(2 w + 3\right)\right]}$