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Question Number 84442 by Power last updated on 13/Mar/20

Answered by john santu last updated on 13/Mar/20

$$\mathrm{let}\mathrm{y}={\mathrm{x}}^{{3\mathrm{x}}^{5\mathrm{x}}}\Rightarrow \ln \left(\mathrm{y}\right)=\ln \left({\mathrm{x}}^{{3\mathrm{x}}^{5\mathrm{x}}}\right)$$$$\mathrm{let}{3\mathrm{x}}^{5\mathrm{x}}=\mathrm{g}\left(\mathrm{x}\right)$$$$\Rightarrow \ln \left(\mathrm{y}\right)=\ln \left({\mathrm{x}}^{\mathrm{g}\left(\mathrm{x}\right)}\right)$$$$\Rightarrow \ln \left(\mathrm{y}\right)=\mathrm{g}\left(\mathrm{x}\right)\ln \left(\mathrm{x}\right)$$$$\Rightarrow \frac{\mathrm{y}{}^{\prime }}{\mathrm{y}}=\frac{\mathrm{g}\left(\mathrm{x}\right)}{\mathrm{x}}+\mathrm{g}{}^{\prime }\left(\mathrm{x}\right)\ln \left(\mathrm{x}\right)$$$$\mathrm{now}\mathrm{consider}\mathrm{g}\left(\mathrm{x}\right)={3\mathrm{x}}^{5\mathrm{x}}$$$$\ln \left(\mathrm{g}\left(\mathrm{x}\right)\right)=\ln \left({3\mathrm{x}}^{5\mathrm{x}}\right)=\ln \left(3\right)+5\mathrm{x}\ln \left(\mathrm{x}\right)$$$$\frac{\mathrm{g}{}^{\prime }\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=0+\frac{5\mathrm{x}}{\mathrm{x}}+5\ln \left(\mathrm{x}\right)$$$$\frac{\mathrm{g}{}^{\prime }\left(\mathrm{x}\right)}{\mathrm{g}\left(\mathrm{x}\right)}=5\ln \left(\mathrm{ex}\right)=\ln {\left(\mathrm{ex}\right)}^5$$$$\mathrm{g}{}^{\prime }\left(\mathrm{x}\right)={15\mathrm{x}}^{5\mathrm{x}}\ln \left(\mathrm{ex}\right)$$$$\mathrm{now}{\mathrm{y}}^{\prime }=\mathrm{y}[{3\mathrm{x}}^{5\mathrm{x}-1}+15\mathrm{xln}\left(\mathrm{ex}\right)\ln \left(\mathrm{x}\right)]$$$$\mathrm{y}{}^{\prime }={\mathrm{x}}^{{3\mathrm{x}}^{5\mathrm{x}}}[{3\mathrm{x}}^{5\mathrm{x}-1}+15\mathrm{xln}\left(\mathrm{ex}\right)\ln \left(\mathrm{x}\right)]$$

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