(x+2)^y=y^x+2Longrightarrow y'(-1)= ?

To find the derivative of \( y \) with respect to \( x \) at the point \( x = -1 \), we need to implicitly differentiate the given equation \( (x+2)^y = y^{x+2} \) with respect to \( x \).<br /><br />Let's differentiate both sides of the equation with respect to \( x \):<br /><br />\[ \frac{d}{dx}[(x+2)^y] = \frac{d}{dx}[y^{x+2}] \]<br /><br />Using the chain rule on the left side, we get:<br /><br />\[ y(x+2)^{y-1} \cdot \frac{d}{dx}(x+2) = \frac{d}{dx}[y^{x+2}] \]<br /><br />\[ y(x+2)^{y-1} \cdot 1 = \frac{d}{dx}[y^{x+2}] \]<br /><br />Now, let's differentiate the right side using the chain rule:<br /><br />\[ \frac{d}{dx}[y^{x+2}] = y^{x+2} \cdot \frac{d}{dx}(x+2) + (x+2) \cdot \frac{d}{dx}[y^{x+2}] \]<br /><br />\[ \frac{d}{dx}[y^{x+2}] = y^{x+2} \cdot 1 + (x+2) \cdot y^{x+2} \cdot \ln(y) \cdot \frac{dy}{dx} \]<br /><br />Now, we can equate the derivatives from both sides:<br /><br />\[ y(x+2)^{y-1} = y^{x+2} + (x+2) \cdot y^{x+2} \cdot \ln(y) \cdot \frac{dy}{dx} \]<br /><br />Now, we need to solve for \( \frac{dy}{dx} \) at \( x = -1 \). Substitute \( x = -1 \) into the equation:<br /><br />\[ y(-1+2)^{y-1} = y^{-1+2} + (-1+2) \cdot y^{-1+2} \cdot \ln(y) \cdot \frac{dy}{dx} \]<br /><br />\[ y \cdot 1^{y-1} = y^1 + 1 \cdot y^1 \cdot \ln(y) \cdot \frac{dy}{dx} \]<br /><br />\[ y = y + y \cdot \ln(y) \cdot \frac{dy}{dx} \]<br /><br />Now, isolate \( \frac{dy}{dx} \):<br /><br />\[ y \cdot \ln(y) \cdot \frac{dy}{dx} = 0 \]<br /><br />Since \( y \neq 0 \), we can divide both sides by \( y \cdot \ln(y) \):<br /><br />\[ \frac{dy}{dx} = 0 \]<br /><br />Therefore, the derivative \( y'(-1) \) is equal to 0.