Let y = x^(4x). If we take the logarithm on both sides of the equation, we get the following:
[tex]\begin{gathered} \ln y=\ln(x^{4x})=4x\ln x \\ \Rightarrow\ln y=4x\ln x \end{gathered}[/tex]
applying implicit derivation, we can find the derivative y':
[tex]\begin{gathered} (\ln y)^{\prime}=(4x\ln x)^{\prime} \\ \Rightarrow\frac{y^{\prime}}{y}=4\ln x+\frac{4x}{x} \\ \Rightarrow y^{\prime}=y(4(\ln x+1)) \\ \Rightarrow y^{\prime}=4x^{4x}(\ln x+1) \end{gathered}[/tex]
therefore, f'(x) = 4x^(4x)(lnx + 1).
Then, for f'(1), we can make x = 1 on the derivative to get the following:
[tex]f^{\prime}(1)=4(1)^{4(1)}(\ln(1)+1)=4(0+1)=4[/tex]
therefore f'(1) = 4
