`y = log_10((x^2-1)/x)` Find the derivative of the function

Thursday, October 31, 2013

$\displaystyle{y}={\log}_{{10}}{\left(\frac{{{{x}}^{{2}}-{1}}}{{x}}\right)}$ Find the derivative of the function

$\displaystyle{y}={\log}_{{10}}{\left(\frac{{{{x}}^{{2}}-{1}}}{{x}}\right)}$

The derivative formula of a logarithm is

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\log}_{{a}}{\left({u}\right)}\right]}=\frac{{1}}{{{\ln{{\left({a}\right)}}}\cdot{u}}}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

Applying this formula, the derivative of the function will be

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{d}}{{\left.{d}{x}}}{\left[{\log}_{{10}}{\left(\frac{{{{x}}^{{2}}-{1}}}{{x}}\right)}\right]}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{1}}{{{\ln{{\left({10}\right)}}}\cdot{\left(\frac{{{{x}}^{{2}}-{1}}}{{x}}\right)}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left(\frac{{{{x}}^{{2}}-{1}}}{{x}}\right)}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{x}}{{{\left({{x}}^{{2}}-{1}\right)}{\ln{{\left({10}\right)}}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left(\frac{{{{x}}^{{2}}-{1}}}{{x}}\right)}$

To get the derivative of $\displaystyle\frac{{{{x}}^{{2}}-{1}}}{{x}}$ , apply quotient rule $\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left(\frac{{{f{{\left({x}\right)}}}}}{{{g{{\left({x}\right)}}}}}\right)}=\frac{{{g{{\left({x}\right)}}}\cdot{f{'}}{\left({x}\right)}-{f{{\left({x}\right)}}}\cdot{g{'}}{\left({x}\right)}}}{{{\left[{g{{\left({x}\right)}}}\right]}}^{{2}}}$ .

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{x}}{{{\left({{x}}^{{2}}-{1}\right)}{\ln{{\left({10}\right)}}}}}\cdot\frac{{{x}\cdot{2}{x}-{\left({{x}}^{{2}}-{1}\right)}\cdot{1}}}{{{x}}^{{2}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{x}}{{{\left({{x}}^{{2}}-{1}\right)}{\ln{{\left({10}\right)}}}}}\cdot\frac{{{2}{{x}}^{{2}}-{{x}}^{{2}}+{1}}}{{{x}}^{{2}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{x}}{{{\left({{x}}^{{2}}-{1}\right)}{\ln{{\left({10}\right)}}}}}\cdot\frac{{{{x}}^{{2}}+{1}}}{{{x}}^{{2}}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{1}}{{{\left({{x}}^{{2}}-{1}\right)}{\ln{{\left({10}\right)}}}}}\cdot\frac{{{{x}}^{{2}}+{1}}}{{x}}$

$\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{{{x}}^{{2}}+{1}}}{{{x}{\left({{x}}^{{2}}-{1}\right)}{\ln{{\left({10}\right)}}}}}$

Therefore, the derivative of the function is $\displaystyle\frac{{{\left.{d}{y}\right.}}}{{\left.{d}{x}}}=\frac{{{{x}}^{{2}}+{1}}}{{{x}{\left({{x}}^{{2}}-{1}\right)}{\ln{{\left({10}\right)}}}}}$ .

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Thursday, October 31, 2013

$\displaystyle{y}={\log}_{{10}}{\left(\frac{{{{x}}^{{2}}-{1}}}{{x}}\right)}$ Find the derivative of the function

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Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

Thursday, October 31, 2013

Find the derivative of the function

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Post a Comment

Thomas Jefferson&#39;s election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?

$\displaystyle{y}={\log}_{{10}}{\left(\frac{{{{x}}^{{2}}-{1}}}{{x}}\right)}$ Find the derivative of the function

Thomas Jefferson's election in 1800 is sometimes called the Revolution of 1800. Why could it be described in this way?