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Question Number 134297 by mathlove last updated on 02/Mar/21

Answered by Ñï= last updated on 02/Mar/21

$$\frac{d}{dx}ln(x^x+2^{x^x})=\frac{1}{x^x+2^{x^x}}[x^x(lnx+1)+\left(ln2\right)2^{x^x}ln{2}^x{2}^{x^x}]$$$$=\frac{1}{x^x+2^{x^x}}[(lnx+1)x^x+{\left(ln2\right)}^2{\left(2^{x^x}\right)}^{2}x]$$$$\frac{d}{dx}\left({lnx}^{ln(x^x+2^{x^x})}\right)$$$$=\frac{d}{dx}\left[\left(lnx\right)ln(x^x+2^{x^x})\right]$$$$=\frac{1}{x}ln(x^x+2^{x^x})+\frac{lnx}{x^x+2^{x^x}}[(lnx+1)x^x+{\left(ln2\right)}^2{\left(2^{x^x}\right)}^{2}x]$$

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