A) \[\sin u\]
B) \[\text{cosec}\,u\]
C) \[2\tan u\]
D) \[3\tan u\]
Correct Answer: D
Solution :
\[(x+y)\,\sin u={{x}^{2}}{{y}^{2}}\] ?(i) On differentiating partially w.r.t. \[x\]of Eq. (i), we get \[(1+0)\sin u+(x+y)\cos u\frac{\partial u}{\partial x}=2x{{y}^{2}}\] \[\Rightarrow \] \[x\sin u+({{x}^{2}}+xy)cosu\frac{\partial u}{\partial x}=2{{x}^{2}}{{y}^{2}}\] ?(ii) On differentiating partially w.r.t. \[y\]of Eq. (i), we get \[(0+1)\sin u+(x+y)\cos u\frac{\partial u}{\partial y}=2{{x}^{2}}y\] \[\Rightarrow \] \[y\sin u+(xy+{{y}^{2}})\cos u\frac{\partial u}{\partial y}=2{{x}^{2}}{{y}^{2}}\] ?(iii) On adding Eqs. (ii) and (iii), we get \[(x+y)\sin u+(x+y)\left\{ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y} \right\}\cos u\] \[=4{{x}^{2}}{{y}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}{{y}^{2}}+(x+y)\left\{ x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y} \right\}\cos u=4{{x}^{2}}{{y}^{2}}\] \[\Rightarrow \] \[(x+y)\left( x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y} \right)\cos u=3{{x}^{2}}{{y}^{2}}\] \[\Rightarrow \] \[(x+y\left( x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y} \right)\cos u\]\[=3(x+y)\sin u\] \[\Rightarrow \] \[x\frac{\partial u}{dx}+y\frac{\partial u}{dy}=3\,\tan u\]
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