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Question Number 13544 by aishadenge last updated on 20/May/17

Commented by prakash jain last updated on 20/May/17

$$\mathrm{What}\mathrm{is}z?$$

Commented by aishadenge last updated on 20/May/17

$$thetoppartisequatedtoznotx.sorry$$$$fortheerror.$$

Commented by prakash jain last updated on 21/May/17

$$\frac{\partial z}{\partial x}=e^{-x}\frac{\partial }{\partial x}(y\cos x+y\sin y)$$$$+(y\cos x+y\sin y)\frac{\partial }{\partial x}{e}^{-x}$$$$=e^{-x}(-y\sin x)-e^{-x}(y\cos x+y\sin y)$$$$=-{ye}^{-x}(\cos x+\sin x+\sin y)$$$$\frac{{\partial }^{2}z}{\partial x^2}=-{ye}^{-x}\frac{\partial }{\partial x}(\cos x+\sin x+\sin y)$$$$-y(\cos x+\sin x+\sin y)\frac{\partial }{\partial x}{e}^{-x}$$$$=-{ye}^{-x}(-\sin x+\cos x)+{ye}^{-x}(\cos x+\sin x+\sin y)$$$$={ye}^{-x}(\sin y+2\sin x)$$$$$$$$\frac{\partial z}{\partial y}=e^{-x}(\cos x+\sin y+y\cos y)$$$$\frac{{\partial }^{2}z}{\partial y^2}=e^{-x}(\cos y+\cos y-y\sin y)$$$$\frac{{\partial }^{2}z}{\partial x^2}+\frac{{\partial }^{2}z}{\partial y^2}\ne 0$$$$\mathrm{Please}\mathrm{recheck}\mathrm{question}.$$$$\mathrm{I}\mathrm{will}\mathrm{also}\mathrm{recheck}\mathrm{answer}.$$

Commented by mrW1 last updated on 21/May/17

$$maybe\mathit{z}={\mathit{e}}^{-\mathit{x}}(\mathit{y}\mathbf{cos}\mathit{x}+\mathit{x}\mathbf{sin}\mathit{y})?$$

Commented by aishadenge last updated on 21/May/17

$$thankyou.$$

Commented by aishadenge last updated on 21/May/17

$$ichecked.itis\mathit{y}\mathit{c}\mathit{o}\mathit{s}\mathit{x}+\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathit{y}indeed$$

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