Implicit Function 90C70D

1. **Problem Statement:**\nWe have two functions:\n(b) $$f(x,y) = xy - \log x + \log y$$ with point $$\left(x_0,y_0\right) = \left(\sqrt{e}, \frac{1}{\sqrt{e}}\right)$$\n(c) $$f(x,y) = y(e^y + x) - \log x$$ with point $$\left(x_0,y_0\right) = (1,0)$$\n\nWe want to verify the hypotheses of the Implicit Function Theorem at these points and then find $$\phi'(x)$$ and $$\phi''(x)$$ where $$y = \phi(x)$$ satisfies $$f(x,\phi(x))=0$$ and $$\phi(x_0) = y_0$$.\n\n---\n\n### Part (b)\n\n2. **Check hypotheses of the Implicit Function Theorem:**\n- Compute partial derivatives:\n$$f_x = y - \frac{1}{x}$$\n$$f_y = x + \frac{1}{y}$$\n- Evaluate at $$\left(\sqrt{e}, \frac{1}{\sqrt{e}}\right)$$:\n$$f_x\left(\sqrt{e}, \frac{1}{\sqrt{e}}\right) = \frac{1}{\sqrt{e}} - \frac{1}{\sqrt{e}} = 0$$\n$$f_y\left(\sqrt{e}, \frac{1}{\sqrt{e}}\right) = \sqrt{e} + \sqrt{e} = 2\sqrt{e} \neq 0$$\n- Since $$f_y \neq 0$$ at the point, the Implicit Function Theorem applies and there exists a function $$\phi(x)$$ near $$x_0$$ such that $$f(x,\phi(x))=0$$ and $$\phi(x_0) = y_0$$.\n\n3. **Find $$\phi'(x)$$:**\nFrom implicit differentiation:\n$$f_x + f_y \phi'(x) = 0 \implies \phi'(x) = -\frac{f_x}{f_y}$$\nAt general $$(x,y)$$:\n$$\phi'(x) = -\frac{y - \frac{1}{x}}{x + \frac{1}{y}} = -\frac{y - \frac{1}{x}}{x + \frac{1}{y}}$$\n\n4. **Evaluate $$\phi'(x)$$ at $$x_0$$:**\n$$\phi'(\sqrt{e}) = -\frac{\frac{1}{\sqrt{e}} - \frac{1}{\sqrt{e}}}{\sqrt{e} + \sqrt{e}} = -\frac{0}{2\sqrt{e}} = 0$$\n\n5. **Find $$\phi''(x)$$:**\nDifferentiate $$\phi'(x)$$ implicitly:\nLet $$u = y - \frac{1}{x}$$ and $$v = x + \frac{1}{y}$$, so $$\phi'(x) = -\frac{u}{v}$$.\n\nUsing quotient rule:\n$$\phi''(x) = -\frac{u'v - uv'}{v^2}$$\n\nCalculate $$u'$$ and $$v'$$:\n$$u' = \frac{dy}{dx} - \left(-\frac{1}{x^2}\right) = \phi'(x) + \frac{1}{x^2}$$\n$$v' = 1 - \frac{1}{y^2} \phi'(x)$$\n\nSubstitute back:\n$$\phi''(x) = -\frac{(\phi'(x) + \frac{1}{x^2})(x + \frac{1}{y}) - (y - \frac{1}{x})(1 - \frac{1}{y^2} \phi'(x))}{\left(x + \frac{1}{y}\right)^2}$$\n\n6. **Evaluate $$\phi''(x)$$ at $$x_0$$:**\nAt $$x_0 = \sqrt{e}$$, $$y_0 = \frac{1}{\sqrt{e}}$$, and $$\phi'(x_0) = 0$$:\nCalculate numerator terms:\n$$u' = 0 + \frac{1}{(\sqrt{e})^2} = \frac{1}{e}$$\n$$v = \sqrt{e} + \sqrt{e} = 2\sqrt{e}$$\n$$u = \frac{1}{\sqrt{e}} - \frac{1}{\sqrt{e}} = 0$$\n$$v' = 1 - \frac{1}{(\frac{1}{\sqrt{e}})^2} \cdot 0 = 1$$\n\nSo numerator:\n$$u'v - uv' = \frac{1}{e} \cdot 2\sqrt{e} - 0 \cdot 1 = \frac{2\sqrt{e}}{e} = 2 e^{-\frac{1}{2}}$$\nDenominator:\n$$v^2 = (2\sqrt{e})^2 = 4e$$\n\nTherefore:\n$$\phi''(\sqrt{e}) = -\frac{2 e^{-\frac{1}{2}}}{4e} = -\frac{2}{4 e^{\frac{3}{2}}} = -\frac{1}{2 e^{\frac{3}{2}}}$$\n\n---\n\n### Part (c)\n\n7. **Check hypotheses of the Implicit Function Theorem:**\n- Compute partial derivatives:\n$$f_x = -\frac{1}{x} + y$$\n$$f_y = e^y + y e^y + x$$\n- Evaluate at $$(1,0)$$:\n$$f_x(1,0) = -1 + 0 = -1 \neq 0$$\n$$f_y(1,0) = e^0 + 0 + 1 = 1 + 0 + 1 = 2 \neq 0$$\n- Since $$f_y \neq 0$$, the Implicit Function Theorem applies.\n\n8. **Find $$\phi'(x)$$:**\n$$\phi'(x) = -\frac{f_x}{f_y} = -\frac{-\frac{1}{x} + y}{e^y + y e^y + x} = \frac{\frac{1}{x} - y}{e^y + y e^y + x}$$\n\n9. **Evaluate $$\phi'(x)$$ at $$x_0=1$$:**\n$$\phi'(1) = \frac{1 - 0}{1 + 0 + 1} = \frac{1}{2}$$\n\n10. **Find $$\phi''(x)$$:**\nLet $$u = \frac{1}{x} - y$$ and $$v = e^y + y e^y + x$$, so $$\phi'(x) = \frac{u}{v}$$\n\nUsing quotient rule:\n$$\phi''(x) = \frac{u'v - uv'}{v^2}$$\n\nCalculate $$u'$$ and $$v'$$:\n$$u' = -\frac{1}{x^2} - \phi'(x)$$\n$$v' = (e^y + y e^y) \phi'(x) + (e^y + y e^y) + 1$$\nNote: derivative of $$e^y + y e^y = e^y + e^y + y e^y = (2 + y) e^y$$, but since $$y$$ depends on $$x$$, chain rule applies:\n$$\frac{d}{dx}(e^y + y e^y) = (2 + y) e^y \phi'(x)$$\n\nSo:\n$$v' = (2 + y) e^y \phi'(x) + 1$$\n\n11. **Evaluate $$\phi''(x)$$ at $$(1,0)$$:**\nAt $$(1,0)$$, $$\phi'(1) = \frac{1}{2}$$, $$u = 1 - 0 = 1$$, $$v = 1 + 0 + 1 = 2$$\n\nCalculate $$u'$$:\n$$u' = -1 - \frac{1}{2} = -\frac{3}{2}$$\nCalculate $$v'$$:\n$$v' = (2 + 0) \cdot 1 \cdot \frac{1}{2} + 1 = 2 \cdot \frac{1}{2} + 1 = 1 + 1 = 2$$\n\nCalculate numerator:\n$$u'v - uv' = \left(-\frac{3}{2}\right) \cdot 2 - 1 \cdot 2 = -3 - 2 = -5$$\nDenominator:\n$$v^2 = 2^2 = 4$$\n\nTherefore:\n$$\phi''(1) = \frac{-5}{4} = -\frac{5}{4}$$\n\n---\n\n**Final answers:**\n- For (b): $$\phi'(\sqrt{e}) = 0$$, $$\phi''(\sqrt{e}) = -\frac{1}{2 e^{\frac{3}{2}}}$$\n- For (c): $$\phi'(1) = \frac{1}{2}$$, $$\phi''(1) = -\frac{5}{4}$$