Question

$$\Delta_{w_{u}}=\alpha(u_{j}-y_{j})y_{j}(1-y_{j})y_{i}$$

Answer

$$t=(IM*a*p*h*y^2*j^2*(u-y)*(1-y*j))/(De*w*u)$$

Solution


Cancel \(l\) on both sides.
\[Detawu=apha(uj-yj)yj(1-yj)y\imath \]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[Detawu={a}^{2}ph(uj-yj){y}^{2}j(1-yj)\imath \]
Regroup terms.
\[Detawu=\imath {a}^{2}ph{y}^{2}j(uj-yj)(1-yj)\]
Divide both sides by \(De\).
\[tawu=\frac{\imath {a}^{2}ph{y}^{2}j(uj-yj)(1-yj)}{De}\]
Factor out the common term \(j\).
\[tawu=\frac{\imath {a}^{2}ph{y}^{2}jj(u-y)(1-yj)}{De}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[tawu=\frac{\imath {a}^{2}ph{y}^{2}{j}^{2}(u-y)(1-yj)}{De}\]
Divide both sides by \(a\).
\[twu=\frac{\frac{\imath {a}^{2}ph{y}^{2}{j}^{2}(u-y)(1-yj)}{De}}{a}\]
Simplify  \(\frac{\frac{\imath {a}^{2}ph{y}^{2}{j}^{2}(u-y)(1-yj)}{De}}{a}\)  to  \(\frac{\imath {a}^{2}ph{y}^{2}{j}^{2}(u-y)(1-yj)}{Dea}\).
\[twu=\frac{\imath {a}^{2}ph{y}^{2}{j}^{2}(u-y)(1-yj)}{Dea}\]
Simplify  \(\frac{\imath {a}^{2}ph{y}^{2}{j}^{2}(u-y)(1-yj)}{Dea}\)  to  \(\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{De}\).
\[twu=\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{De}\]
Divide both sides by \(w\).
\[tu=\frac{\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{De}}{w}\]
Simplify  \(\frac{\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{De}}{w}\)  to  \(\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{Dew}\).
\[tu=\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{Dew}\]
Divide both sides by \(u\).
\[t=\frac{\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{Dew}}{u}\]
Simplify  \(\frac{\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{Dew}}{u}\)  to  \(\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{Dewu}\).
\[t=\frac{\imath aph{y}^{2}{j}^{2}(u-y)(1-yj)}{Dewu}\]