Differentiate the following functions:

f(x) = ln ((x^5(3x-5)^2)/((x+1)^12(1-2x)^7)

g(x) = (e^x-ln(x))^tanx

Community Answer

【General guidance】The answer provided below has been developed in a clear step by step manner.Step1/2) \( \mathrm{{f{{\left({x}\right)}}}={\ln{{\left(\frac{{{x}^{{5}}{\left({3}{x}-{5}\right)}^{{2}}}}{{{\left({x}+{1}\right)}^{{12}}{\left({1}-{2}{x}\right)}^{{7}}}}\right)}}}} \)Using property of logarthim \( \mathrm{{{\ln{{x}}}^{{n}}=}{n}{\ln{{x}}},{\ln{{\left(\frac{{m}}{{n}}\right)}}}={\ln{{m}}}-{\ln{{n}}}} \)\( \mathrm{{f{{\left({x}\right)}}}={5}{\ln{{x}}}+{2}{\ln{{\left({3}{x}-{5}\right)}}}-{12}{\ln{{\left({x}+{1}\right)}}}-{7}{\ln{{\left({1}-{2}{x}\right)}}}} \)By the Sum Rule, the derivative of \( \mathrm{{5}{\ln{{\left({x}\right)}}}+{2}{\ln{{\left({3}{x}-{5}\right)}}}-{12}{\ln{{\left({x}+{1}\right)}}}-{7}{\ln{{\left({1}-{2}{x}\right)}}}} \) with respect to \( \mathrm{{x}} \) is \( \mathrm{\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[{5}{\ln{{\left({x}\right)}}}\right]}+\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[{2}{\ln{{\left({3}{x}-{5}\right)}}}\right]}+\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[-{12}{\ln{{\left({x}+{1}\right)}}}\right]}+\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[-{7}{\ln{{\left({1}-{2}{x}\right)}}}\right]}} \).\( \mathrm{\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[{5}{\ln{{\left({x}\right)}}}\right]}+\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[{2}{\ln{{\left({3}{x}-{5}\right)}}}\right]}+\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[-{12}{\ln{{\left({x}+{1}\right)}}}\right]}+\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[-{7}{\ln{{\left({1}-{2}{x}\right)}}}\right]}} \)Evaluate \( \mathrm{\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[{5}{\ln{{\left({x}\right)}}}\right]}} \).\( \mathrm{\frac{{{5}}}{{{x}}}+\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[{2}{\ln{{\left({3}{x}-{5}\right)}}}\right]}+\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[-{12}{\ln{{\left({x}+{1}\right)}}}\right]}+\frac{{{d}}}{{{\left.{d}{x}\right.}}}{\left[-{7}{\ln{{\left({1}-{2}{x}\right)}}}\right]}} \)Evaluate \( \mathr ... See the full answer