In Exercises 1–24, find the derivative of y with respect to the appropriate variable.
13. y = (x+2)^(x+2)

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Recognize that the function is of the form \(y = f(x)^{g(x)}\), where both the base and the exponent depend on \(x\). This suggests using logarithmic differentiation.

Take the natural logarithm of both sides: \(\ln y = \ln \left( (x+2)^{x+2} \right)\).

Use the logarithm power rule to simplify the right side: \(\ln y = (x+2) \cdot \ln (x+2)\).

Differentiate both sides with respect to \(x\). For the left side, use implicit differentiation: \(\frac{1}{y} \frac{dy}{dx}\). For the right side, apply the product rule to \((x+2) \cdot \ln (x+2)\).

After differentiating, solve for \(\frac{dy}{dx}\) by multiplying both sides by \(y\), and then substitute back \(y = (x+2)^{x+2}\) to express the derivative in terms of \(x\).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Implicit Differentiation and Logarithmic Differentiation

When a function has a variable in both the base and the exponent, such as y = (x+2)^(x+2), logarithmic differentiation is used. Taking the natural logarithm of both sides simplifies the expression, allowing differentiation of the exponent and base separately.

Derivative of Exponential Functions with Variable Exponents

For functions where the exponent is a variable, the derivative involves applying the chain rule and product rule after rewriting the function using logarithms. This approach helps handle the complexity of differentiating expressions like a(x)^{b(x)}.

Chain Rule and Product Rule

The chain rule is used to differentiate composite functions, while the product rule applies when differentiating products of functions. Both are essential here because the function involves a product of terms after logarithmic transformation.

In Exercises 1–24, find the derivative of y with respect to the appropriate variable.13. y = (x+2)^(x+2)