Differentiability MCQ Quiz in తెలుగు - Objective Question with Answer for Differentiability - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

$\frac{\partial \mathrm{u}}{\partial \mathrm{z}}=\mathrm{x}\mathrm{y}{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}$

$\frac{\partial }{\partial \mathrm{y}}\left(\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right)=\frac{\partial }{\partial \mathrm{y}}(\mathrm{x}\mathrm{y}{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}})$

$\frac{{\partial }^2\mathrm{u}}{\partial \mathrm{y}\partial \mathrm{z}}=\mathrm{x}\mathrm{y}\frac{\partial }{\partial \mathrm{y}}({\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}})+{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}\frac{\partial }{\partial \mathrm{y}}(\mathrm{x}\mathrm{y})$

= xy(xz)e^xyz + xe^xyz

$\frac{\partial }{\partial \mathrm{x}}\left(\frac{{\partial }^2\mathrm{u}}{\partial \mathrm{y}\partial \mathrm{z}}\right)=\frac{\partial }{\partial \mathrm{x}}({\mathrm{x}}^2\mathrm{y}\mathrm{z}+\mathrm{x}){\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}$

$\frac{{\partial }^3\mathrm{u}}{\partial \mathrm{x}\partial \mathrm{y}\partial \mathrm{z}}=({\mathrm{x}}^2\mathrm{y}\mathrm{z}+\mathrm{x})\mathrm{y}\mathrm{z}{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}+{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}(2\mathrm{x}\mathrm{y}\mathrm{z}+1)$

$={\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}({\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2+\mathrm{x}\mathrm{y}\mathrm{z}+2\mathrm{x}\mathrm{y}\mathrm{z}+1)$

= (1 + 3xyz + x²y²z²) e^xyz

x, y, z = 1, 1, 1 रखने पर हमें प्राप्त होता है

$\frac{{\partial }^3\mathrm{u}}{\partial \mathrm{x}\partial \mathrm{y}\partial \mathrm{z}}=(1+3+1)\mathrm{e}$

= 5e

$\frac{\partial \mathrm{u}}{\partial \mathrm{z}}=\mathrm{x}\mathrm{y}{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}$

$\frac{\partial }{\partial \mathrm{y}}\left(\frac{\partial \mathrm{u}}{\partial \mathrm{z}}\right)=\frac{\partial }{\partial \mathrm{y}}(\mathrm{x}\mathrm{y}{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}})$

$\frac{{\partial }^2\mathrm{u}}{\partial \mathrm{y}\partial \mathrm{z}}=\mathrm{x}\mathrm{y}\frac{\partial }{\partial \mathrm{y}}({\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}})+{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}\frac{\partial }{\partial \mathrm{y}}(\mathrm{x}\mathrm{y})$

= xy(xz)e^xyz + xe^xyz

$\frac{\partial }{\partial \mathrm{x}}\left(\frac{{\partial }^2\mathrm{u}}{\partial \mathrm{y}\partial \mathrm{z}}\right)=\frac{\partial }{\partial \mathrm{x}}({\mathrm{x}}^2\mathrm{y}\mathrm{z}+\mathrm{x}){\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}$

$\frac{{\partial }^3\mathrm{u}}{\partial \mathrm{x}\partial \mathrm{y}\partial \mathrm{z}}=({\mathrm{x}}^2\mathrm{y}\mathrm{z}+\mathrm{x})\mathrm{y}\mathrm{z}{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}+{\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}(2\mathrm{x}\mathrm{y}\mathrm{z}+1)$

$={\mathrm{e}}^{\mathrm{x}\mathrm{y}\mathrm{z}}({\mathrm{x}}^2{\mathrm{y}}^2{\mathrm{z}}^2+\mathrm{x}\mathrm{y}\mathrm{z}+2\mathrm{x}\mathrm{y}\mathrm{z}+1)$

= (1 + 3xyz + x²y²z²) e^xyz

x, y, z = 1, 1, 1 रखने पर हमें प्राप्त होता है

$\frac{{\partial }^3\mathrm{u}}{\partial \mathrm{x}\partial \mathrm{y}\partial \mathrm{z}}=(1+3+1)\mathrm{e}$

= 5e