y=f(x)=(x^x)^x=x^x^2 lny=x^2 lnx (1)/(y) y^'=x+2 x lnx y^'=(x^x)^x(x+2 x lnx)

step 1 In this question we are given with the function that is y equals to 10 inverse 2 raise to the po

y=f(x)=(x^x)^x=x^x^2 lny=x^2 lnx (1)/(y)...

$\begin{aligned} &y=f(x)=\left(x^{x}\right)^{x}=x^{x^{2}} \\ &\ln y=x^{2} \ln x \\ &\frac{1}{y} y^{\prime}=x+2 x \ln x \\ &y^{\prime}=\left(x^{x}\right)^{x}(x+2 x \ln x) \\ &f^{\prime}(1)=1 \\ &\text { Now, } y=g(x)=x^{\left(x^{-}\right)} \\ &\ln y=x^{x} \ln x \\ &\frac{1}{y} y^{\prime}=x^{x-1}+(\ln x)\left(x^{x}\right)(1+\ln x) \\ &y^{\prime}=g^{\prime}(x)=x^{\left(x^{*}\right)}\left(x^{x-1}+x^{x} \ln x(1+\ln x)\right) \\ &g^{\prime}(1)=1 \end{aligned}$

$\begin{aligned}
&y=f(x)=\left(x^{x}\right)^{x}=x^{x^{2}} \\
&\ln y=x^{2} \ln x \\
&\frac{1}{y} y^{\prime}=x+2 x \ln x \\
&y^{\prime}=\left(x^{x}\right)^{x}(x+2 x \ln x) \\
&f^{\prime}(1)=1 \\
&\text { Now, } y=g(x)=x^{\left(x^{-}\right)} \\
&\ln y=x^{x} \ln x \\
&\frac{1}{y} y^{\prime}=x^{x-1}+(\ln x)\left(x^{x}\right)(1+\ln x) \\
&y^{\prime}=g^{\prime}(x)=x^{\left(x^{*}\right)}\left(x^{x-1}+x^{x} \ln x(1+\ln x)\right) \\
&g^{\prime}(1)=1
\end{aligned}$

Key Concepts

Chain Rule

The chain rule is a fundamental differentiation technique used to compute the derivative of composite functions. When a function is composed of several layers (for example, a logarithm of a function), the chain rule allows one to systematically differentiate each layer, ensuring that the effect of each nested function is properly accounted for.

Implicit Differentiation

Implicit differentiation is used when the function is given in an implicit form, where the dependent variable is not isolated on one side of the equation. It involves differentiating both sides of the equation with respect to the independent variable and then solving for the derivative of the dependent variable.

Exponential Functions with Variable Exponents

Exponential functions where the exponent itself is a function of the variable require special handling. In such cases, the standard rules of differentiation are not directly applicable, so the function is often transformed to a form that makes the exponent easier to differentiate, typically by using logarithms.

Logarithmic Differentiation

Logarithmic differentiation is a method used to differentiate functions by taking the natural logarithm of both sides of the equation. This approach is especially useful when dealing with functions that have variables in both the base and the exponent, as it converts multiplicative and exponential relationships into additive ones that are easier to differentiate.

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