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Question Number 215468 by alephnull last updated on 07/Jan/25

$$\frac{\partial x\omega }{\partial y\omega }\cdot \frac{\partial e}{\partial \omega }=?$$$$x=f\left(y\right)$$$$\omega =g\left(y\right)$$$$e=h\left(\omega \right)$$

Answered by MrGaster last updated on 08/Jan/25

$$\frac{\partial x}{\partial y}\cdot \frac{\partial w}{\partial \omega }\cdot \frac{\partial e}{\partial \omega }$$$$=\frac{\partial x}{\partial y}\cdot 1\cdot \frac{\partial e}{\partial \omega }$$$$=\frac{\partial x}{\partial y}\cdot \frac{\partial e}{\partial \omega }$$$$\mathrm{Given}:$$$$x=f\left(y\right)$$$$\omega =g\left(y\right)$$$$e=h\left(\omega \right)$$$$\mathrm{Thus}:$$$$\frac{\partial x}{\partial y}=f^{\prime }\left(y\right)$$$$\frac{\partial e}{\partial \omega }=h^{\prime }\left(\omega \right)$$$$\mathrm{Substitute}\mathrm{back}:$$$$\frac{\partial x\omega }{\partial y\omega }\cdot \frac{\partial e}{\partial \omega }=f^{\prime }\left(y\right)\cdot h^{\prime }\left(g\left(y\right)\right)$$$$\mathrm{Conclusion}:$$$$f^{\prime }\left(y\right)\cdot h^{\prime }\left(g\left(y\right)\right)$$

Commented by alephnull last updated on 08/Jan/25

$$\mathrm{thank}\mathrm{you}$$

Answered by issac last updated on 08/Jan/25

$$=\frac{ex}{y\omega }$$

Commented by MathematicalUser2357 last updated on 29/Mar/25

$$\mathrm{Did}\mathrm{you}\mathrm{miscalculated}\mathrm{because}:$$$$\left(\mathrm{A}\right)\mathrm{you}\mathrm{forgot}\mathrm{the}\mathrm{notation}\mathrm{of}\mathrm{derivatives}$$$$\left(\mathrm{B}\right)\mathrm{you}\mathrm{forgot}\mathrm{the}\mathrm{first}\mathrm{principle}\mathrm{of}\mathrm{derivatives}$$$$\left(\mathrm{C}\right)\mathrm{you}\mathrm{thinked}\mathrm{that}\mathrm{as}\mathrm{an}\mathrm{algebra}\mathrm{question}$$

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