Find the nonpermissible replacement for n in this

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stephstewart1209

13.07.2021

$\ln(x+y) = \dfrac{e^x}{y}$ answer: $\dfrac{dy}{dx} = \dfrac{xye^x+y^2e^x -y^2}{y^2 + xe^x + ye^x}$ ======use implicit differentiation: differentiate both sides with respect to x. $\begin{aligned} \dfrac{d}{dx} \left[ \ln(x+y) \right] & = \dfrac{d}{dx} \left[\dfrac{e^x}{y}\right] \end{aligned}$ focusing on the left-hand side, use the chain rule.the outside "function" is ln(x) and the inside "function" is x+ythe derivative of outside function, ln(x) is 1/x. stick the inside function into the derivative of outside to get 1/(x+y).then multiply by derivative of inside. $\begin{aligned} \dfrac{d}{dx} \left[ \ln(x+y) \right] & = \frac{1}{x+y} \cdot \tfrac{d}{dx}(x+y) \\ & = \frac{\tfrac{d}{dx}(x+y)}{x+y} \\ & = \frac{\tfrac{d}{dx}(x)+\tfrac{d}{dx}(y)}{x+y} \\ & = \frac{1+\tfrac{dy}{dx}}{x+y} \end{aligned}$ focusing on the right-hand side, use the quotient rule."low d high minus high d low; draw a line and square below." $\begin{aligned} \dfrac{d}{dx} \left[\dfrac{e^x}{y}\right] & = \dfrac{y \cdot \tfrac{d}{dx}(e^x) - e^x \cdot \tfrac{d}{dx}(y) }{y^2} \\ & = \dfrac{ye^x - e^x\tfrac{dy}{dx} }{y^2} \end{aligned}$ joining these two together in the original equation: $\begin{aligned} \dfrac{d}{dx} \left[ \ln(x+y) \right] & = \dfrac{d}{dx} \left[\dfrac{e^x}{y}\right] \\ \frac{1+\tfrac{dy}{dx}}{x+y} & = \dfrac{ye^x - e^x\tfrac{dy}{dx} }{y^2} \end{aligned}$ multiply both sides by $y^2$ to cancel out the $y^2$ in the denominator. $\begin{aligned}\frac{y^2\left(1+\tfrac{dy}{dx}\right)}{x+y} & = ye^x - e^x\tfrac{dy}{dx} \end{aligned}$ multiply both sides by $x+y$ to cancel out the $x+y$ in the denominator.then start expanding. we need to solve for dy/dx. $\begin{aligned}y^2\left(1+\tfrac{dy}{dx}\right) & = (x+y)\left(ye^x - e^x\tfrac{dy}{dx}\right) \\ y^2 + y^2 \tfrac{dy}{dx} & = ye^x(x+y) - e^x\tfrac{dy}{dx} (x+y) \\ y^2 + y^2 \tfrac{dy}{dx} & = xye^x+y^2e^x - xe^x\tfrac{dy}{dx}-ye^x\tfrac{dy}{dx} \end{aligned}$ move all terms with $\tfrac{dy}{dx}$ together; move everything else to the other side. $\begin{aligned} y^2 + y^2 \tfrac{dy}{dx} & = xye^x+y^2e^x \boxed{- xe^x\tfrac{dy}{dx}}-ye^x\tfrac{dy}{dx} \\ y^2 + y^2 \tfrac{dy}{dx} + xe^x\tfrac{dy}{dx} & = xye^x+y^2e^x \boxed{-ye^x\tfrac{dy}{dx}} \\ \boxed{y^2} + y^2 \tfrac{dy}{dx} + xe^x\tfrac{dy}{dx} + ye^x\tfrac{dy}{dx} & = xye^x+y^2e^x \\ y^2 \tfrac{dy}{dx} + xe^x\tfrac{dy}{dx} + ye^x\tfrac{dy}{dx} & = xye^x+y^2e^x -y^2 \end{aligned}$ factor out dy/dx from left-side, then divide both sides by that large thing to isolate dy/dx. $\begin{aligned} \tfrac{dy}{dx}\left(y^2 + xe^x + ye^x\right) & = xye^x+y^2e^x -y^2 \\ \frac{dy}{dx} & = \dfrac{xye^x+y^2e^x -y^2}{y^2 + xe^x + ye^x} \end{aligned}$

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Find the nonpermissible replacement for n in this expression n+2/n+8

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