2 (x+2)^y=y^x+2Longrightarrow y'(-1) 3. (tanx)^coty=y^e^(x)Longrightarrow y' ti

Let's correct the approach and solve the problem step by step.<br /><br />Given the equation:<br />\[ 2 (x+2)^{y} = y^{x+2} \]<br /><br />We need to find \( y'(-1) \).<br /><br />First, take the natural logarithm (ln) of both sides to make the equation easier to differentiate:<br />\[ \ln \left( 2 (x+2)^{y} \right) = \ln \left( y^{x+2} \right) \]<br /><br />Using the properties of logarithms, we get:<br />\[ \ln(2) + \ln((x+2)^y) = \ln(y^{x+2}) \]<br />\[ \ln(2) + y \ln(x+2) = (x+2) \ln(y) \]<br /><br />Now, differentiate both sides with respect to \( x \):<br />\[ \frac{d}{dx} \left[ \ln(2) + y \ln(x+2) \right] = \frac{d}{dx} \left[ (x+2) \ln(y) \right] \]<br /><br />The derivative of a constant is zero, so:<br />\[ y \cdot \frac{1}{x+2} \cdot \frac{d}{dx}(x+2) + y' \ln(x+2) = \ln(y) \cdot \frac{d}{dx}(x+2) + (x+2) \cdot \frac{1}{y} \cdot y' \]<br /><br />Simplify the derivatives:<br />\[ y \cdot \frac{1}{x+2} + y' \ln(x+2) = \ln(y) \cdot 1 + (x+2) \cdot \frac{1}{y} \cdot y' \]<br /><br />Combine like terms:<br />\[ y' \ln(x+2) - (x+2) \cdot \frac{1}{y} \cdot y' = \ln(y) - \frac{y}{x+2} \]<br /><br />Factor out \( y' \):<br />\[ y' \left( \ln(x+2) - \frac{x+2}{y} \right) = \ln(y) - \frac{y}{x+2} \]<br /><br />Solve for \( y' \):<br />\[ y' = \frac{\ln(y) - \frac{y}{x+2}}{\ln(x+2) - \frac{x+2}{y}} \]<br /><br />Now, substitute \( x = -1 \) into the equation:<br />\[ y'(-1) = \frac{\ln(y) - \frac{y}{-1+2}}{\ln(-1+2) - \frac{-1+2}{y}} \]<br />\[ y'(-1) = \frac{\ln(y) + y}{\ln(1) - \frac{1}{y}} \]<br /><br />Since \( \ln(1) = 0 \):<br />\[ y'(-1) = \frac{\ln(y) + y}{-\frac{1}{y}} \]<br />\[ y'(-1) = -y (\ln(y) + y) \]<br /><br />So, the derivative \( y'(-1) \) is:<br />\[ y'(-1) = -y (\ln(y) + y) \]<br /><br />For the second equation:<br />\[ (\tan(x))^{(\cot(y))} = y^{e^x} \]<br /><br />We need to find \( y' \).<br /><br />Take the natural logarithm of both sides:<br />\[ \ln \left( (\tan(x))^{(\cot(y))} \right) = \ln \left( y^{e^x} \right) \]<br /><br />Using the properties of logarithms:<br />\[ \cot(y) \ln(\tan(x)) = e^x \ln(y) \]<br /><br />Differentiate both sides with respect to \( x \):<br />\[ \frac{d}{dx} \left[ \cot(y) \ln(\tan(x)) \right] = \frac{d}{dx} \left[ e^x \ln(y) \right] \]<br /><br />Using the product rule:<br />\[ \cot(y) \cdot \frac{1}{\tan(x)} \cdot \frac{d}{dx}(\tan(x)) + \ln(\tan(x)) \cdot \frac{d}{dx}(\cot(y)) = e^x \cdot \frac{1}{y} \cdot y' \]<br /><br />Simplify the derivatives:<br />\[ \cot(y) \cdot \sec^2(x) + \ln(\tan(x)) \cdot (-\csc^2(y)) \cdot y' = e^x \cdot \frac{1}{y} \cdot y' \]<br /><br />Combine like terms:<br />\[ \cot(y) \cdot \sec^2(x) - \ln(\tan(x)) \cdot \csc^2(y) \cdot y'