Question Number 108927 by mohammad17 last updated on 20/Aug/20
Commented by mohammad17 last updated on 20/Aug/20
$${help}\:{me}\:{sir} \\ $$
Answered by 1549442205PVT last updated on 21/Aug/20
$$\mathrm{z}=\mathrm{sinxy}\Rightarrow\frac{\mathrm{dz}}{\partial\mathrm{x}}=\mathrm{ycosxy}\left(\mathrm{1}\right) \\ $$$$\mathrm{u}=\mathrm{y}^{\mathrm{2}} −\mathrm{4x}\Rightarrow\frac{\mathrm{du}}{\partial\mathrm{x}}=−\mathrm{4}\Rightarrow\frac{\mathrm{dx}}{\mathrm{du}}=−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}\right) \\ $$$$\mathrm{Multiplying}\:\mathrm{two}\:\mathrm{the}\:\mathrm{equalities}\:\mathrm{above} \\ $$$$\mathrm{we}\:\mathrm{get}\:\frac{\mathrm{dz}}{\mathrm{du}}=\frac{−\mathrm{ycosxy}}{\mathrm{4}}\mid\left(\mathrm{0},\mathrm{1},\mathrm{1}\right)=\frac{−\mathrm{1}.\mathrm{cos}\left(\mathrm{0}.\mathrm{1}\right)}{\mathrm{4}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{4}}=−\mathrm{0}.\mathrm{25}\Rightarrow\mathrm{Choose}\:\mathrm{B} \\ $$
Commented by mohammad17 last updated on 21/Aug/20
$${thank}\:{you}\:{sir} \\ $$