Question Number 49296 by peter frank last updated on 05/Dec/18
Answered by kaivan.ahmadi last updated on 05/Dec/18
$$\frac{\mathrm{df}}{\mathrm{dx}}=\mathrm{2xysiny}^{\mathrm{2}} \mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{ycosy}^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{and} \\ $$$$\frac{\mathrm{df}}{\mathrm{dy}}=\mathrm{x}^{\mathrm{2}} \mathrm{siny}^{\mathrm{2}} \mathrm{x}+\mathrm{x}^{\mathrm{2}} \mathrm{y}.\mathrm{2ycosy}^{\mathrm{2}} \mathrm{x}= \\ $$$$\mathrm{x}^{\mathrm{2}} \mathrm{siny}^{\mathrm{2}} \mathrm{x}+\mathrm{2x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \mathrm{cosy}^{\mathrm{2}} \mathrm{x} \\ $$
Answered by kaivan.ahmadi last updated on 05/Dec/18
$$\mathrm{2}. \\ $$$$\frac{\mathrm{df}}{\mathrm{dx}}=\mathrm{ycosxy} \\ $$$$\frac{\mathrm{df}}{\mathrm{dy}}=\mathrm{xcosxy} \\ $$