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Question Number 105370 by mohammad17 last updated on 28/Jul/20

Answered by Ar Brandon last updated on 28/Jul/20

$$\mathrm{Q}1/\mathrm{A}/$$$$\left(1\right)\mathrm{f}(x,y)=e^{-2y}\cos \left(3x\right)$$$$\frac{\delta \mathrm{f}}{\delta x}=-3e^{-2y}\sin \left(3x\right)\Rightarrow \frac{{\delta }^2\mathrm{f}}{\delta x^2}=-9e^{-2y}\cos \left(3x\right)$$$$\frac{\delta \mathrm{f}}{\delta y}=-2e^{-2y}\cos \left(3x\right)\Rightarrow \frac{{\delta }^2\mathrm{f}}{\delta y^2}=4e^{-2y}\cos \left(3x\right)$$$$\frac{{\delta }^2\mathrm{f}}{\delta x^2}+\frac{{\delta }^2\mathrm{f}}{\delta y^2}=-5e^{-2y}\cos \left(3\mathrm{x}\right)\ne 0$$$$\mathrm{However},\mathrm{if}\mathrm{f}\left(x\right)=e^{-3y}\cos \left(3x\right),{\mathrm{we}}^{\prime }\mathrm{ll}\mathrm{have};$$$$\frac{{\delta }^2\mathrm{f}}{\delta x^2}=-9e^{-3y}\cos \left(3x\right)\mathrm{and}\frac{{\delta }^2\mathrm{f}}{\delta y^2}=9e^{-3y}\cos \left(3x\right)$$$$\Rightarrow \frac{{\delta }^2\mathrm{f}}{\delta x^2}+\frac{{\delta }^2\mathrm{f}}{\delta y^2}=0$$

Commented by mohammad17 last updated on 28/Jul/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 28/Jul/20

$${\int }_0^{log2}{\int }_1^{log4}{e}^{2x+y}dydx$$$${\int }_0^{log2}{e}^{2x}{\left[e^y\right]}_1^{log4}$$$$(4-e){\int }_0^{log2}{e}^{2x}dx$$$$\frac{4-e}{2}{\left[e^{2x}\right]}_0^{log2}=\frac{4-e}{2}(4-1)=\frac{3}{2}(4-e)$$

Commented by mohammad17 last updated on 28/Jul/20

$$thankyousir$$

Answered by Dwaipayan Shikari last updated on 28/Jul/20

$${\int }_1^2{\int }_0^4\frac{\sqrt{x}}{y^3}dxdy$$$$={\int }_1^2\frac{2}{3y^3}{\left[x^{\frac{3}{2}}\right]}_0^{4}dy$$$$={\int }_1^2\frac{16}{3}{y}^{-3}dy$$$$=-{\left[\frac{8}{3y^2}\right]}_1^2=-\frac{2}{3}+\frac{8}{3}=2$$

Commented by mohammad17 last updated on 28/Jul/20

$$thankyousir$$

Answered by Ar Brandon last updated on 28/Jul/20

$$\mathrm{Q}1\mathrm{B}$$$$\mathrm{W}=\ln (x^2+y^2+3z)$$$$\frac{\delta \mathrm{W}}{\delta x}=\frac{2x}{x^2+y^2+3z},\frac{{\delta }^2\mathrm{W}}{\delta x\delta y}=\frac{-4xy}{{(x^2+y^2+3z)}^2}$$$$\frac{{\delta }^3\mathrm{W}}{\delta x\delta y\delta z}=\frac{12xyz}{{(x^2+y^2+3z)}^3}$$

Commented by mohammad17 last updated on 28/Jul/20

$$thankyousir$$

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