F(x,y) = x^2 + y^2 + sin (xy)
Added by Alba A.
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Certainly! If the question is to find the partial derivatives of the function \(F(x,y) = x^2 + y^2 + \sin(xy)\) with respect to \(x\) and \(y\), here's how you would approach it: ### Finding \(\frac{\partial F}{\partial x}\) ** Show more…
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$\frac{f(2 x+2 y)-f(2 x-2 y)}{f(2 x+2 y)+f(2 x-2 y)}=\frac{\cos x \sin y}{\sin x \cos y}$ Applying C \& D, $\frac{\mathrm{f}(2 \mathrm{x}+2 \mathrm{y})}{\mathrm{f}(2 \mathrm{x}-2 \mathrm{y})}=\frac{\sin (\mathrm{x}+\mathrm{y})}{\sin (\mathrm{x}-\mathrm{y})}$ $\frac{\mathrm{f}(2 \mathrm{x}+2 \mathrm{y})}{\sin (\mathrm{x}+\mathrm{y})}$ is constant $\Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{C} \sin \frac{\mathrm{x}}{2}$ $\mathrm{C}=1$ using $\mathrm{f}^{\prime}(0)=\frac{1}{2}$ $\Rightarrow \mathrm{f}(\mathrm{x})=\sin \frac{\mathrm{x}}{2}$ $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2} \cos \frac{\mathrm{x}}{2}$ $\mathrm{f}^{\prime \prime}(\mathrm{x})=\frac{-1}{4} \sin \frac{\mathrm{x}}{2}$ $\Rightarrow 4 \mathrm{f}^{\prime \prime}(\mathrm{x})+\mathrm{f}(\mathrm{x})=0$
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