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x-y-e-x-f-y-g-y-e-h-




Question Number 215468 by alephnull last updated on 07/Jan/25
((∂xω)/(∂yω))∙(∂e/∂ω)=?  x=f(y)  ω=g(y)  e=h(ω)
$$\frac{\partial{x}\omega}{\partial{y}\omega}\centerdot\frac{\partial{e}}{\partial\omega}=? \\ $$$${x}={f}\left({y}\right) \\ $$$$\omega={g}\left({y}\right) \\ $$$${e}={h}\left(\omega\right) \\ $$
Answered by MrGaster last updated on 08/Jan/25
(∂x/∂y)∙(∂w/∂ω)∙(∂e/∂ω)  =(∂x/∂y)∙1∙(∂e/∂ω)  =(∂x/∂y)∙(∂e/∂ω)  Given:  x=f(y)  ω=g(y)  e=h(ω)  Thus:  (∂x/∂y)=f′(y)  (∂e/∂ω)=h′(ω)  Substitute back:  ((∂xω)/(∂yω))∙(∂e/∂ω)=f′(y)∙h′(g(y))  Conclusion:  f′(y)∙h′(g(y))
$$\frac{\partial{x}}{\partial{y}}\centerdot\frac{\partial{w}}{\partial\omega}\centerdot\frac{\partial{e}}{\partial\omega} \\ $$$$=\frac{\partial{x}}{\partial{y}}\centerdot\mathrm{1}\centerdot\frac{\partial{e}}{\partial\omega} \\ $$$$=\frac{\partial{x}}{\partial{y}}\centerdot\frac{\partial{e}}{\partial\omega} \\ $$$$\mathrm{Given}: \\ $$$${x}={f}\left({y}\right) \\ $$$$\omega={g}\left({y}\right) \\ $$$${e}={h}\left(\omega\right) \\ $$$$\mathrm{Thus}: \\ $$$$\frac{\partial{x}}{\partial{y}}={f}'\left({y}\right) \\ $$$$\frac{\partial{e}}{\partial\omega}={h}'\left(\omega\right) \\ $$$$\mathrm{Substitute}\:\mathrm{back}: \\ $$$$\frac{\partial{x}\omega}{\partial{y}\omega}\centerdot\frac{\partial{e}}{\partial\omega}={f}'\left({y}\right)\centerdot{h}'\left({g}\left({y}\right)\right) \\ $$$$\mathrm{Conclusion}: \\ $$$${f}'\left({y}\right)\centerdot{h}'\left({g}\left({y}\right)\right) \\ $$
Commented by alephnull last updated on 08/Jan/25
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by issac last updated on 08/Jan/25
=((ex)/(yω))
$$=\frac{{ex}}{{y}\omega} \\ $$

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