`y = ln(tanh(x/2))` Find the derivative of the function

Saturday, December 11, 2010

$\displaystyle{y}={\ln{{\left({\tanh{{\left(\frac{{x}}{{2}}\right)}}}\right)}}}$ Find the derivative of the function

$\displaystyle{y}={\ln{{\left({\tanh{{\left(\frac{{x}}{{2}}\right)}}}\right)}}}$

The derivative formula of natural logarithm is

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

Applying this formula, the derivative of the function will be

$\displaystyle{y}'=\frac{{d}}{{\left.{d}{x}}}{\left[{\ln{{\left({\tanh{{\left(\frac{{x}}{{2}}\right)}}}\right)}}}\right]}$

$\displaystyle{y}'=\frac{{1}}{{{\tanh{{\left(\frac{{x}}{{2}}\right)}}}}}\cdot\frac{{d}}{{\left.{d}{x}}}{\left[{\tanh{{\left(\frac{{x}}{{2}}\right)}}}\right]}$

To take the derivative of hyperbolic tangent, apply the formula

$\displaystyle\frac{{d}}{{\left.{d}{x}}}{\left[{\tanh{{\left({u}\right)}}}\right]}={{\sec{{h}}}}^{{2}}{\left({u}\right)}\cdot\frac{{{d}{u}}}{{\left.{d}{x}}}$

So y' will become

$\displaystyle{y}'=\frac{{1}}{{{\tanh{{\left(\frac{{x}}{{2}}\right)}}}}}\cdot{{\sec{{h}}}}^{{2}}{\left(\frac{{x}}{{2}}\right)}\cdot\frac{{d}}{{\left.{d}{x}}}{\left(\frac{{x}}{{2}}\right)}$

$\displaystyle{y}'=\frac{{1}}{{{\tanh{{\left(\frac{{x}}{{2}}\right)}}}}}\cdot{{\sec{{h}}}}^{{2}}{\left(\frac{{x}}{{2}}\right)}\cdot\frac{{1}}{{2}}$

$\displaystyle{y}'=\frac{{{{\sec{{h}}}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}{{{2}{\tanh{{\left(\frac{{x}}{{2}}\right)}}}}}$

To simplify it further, express it in terms of hyperbolic sine and hyperbolic cosine.

$\displaystyle{\sec{{h}}}{\left({u}\right)}=\frac{{1}}{{\cosh{{\left({u}\right)}}}}$

$\displaystyle{\tanh{{\left({u}\right)}}}=\frac{{\sinh{{\left({u}\right)}}}}{{\cosh{{\left({u}\right)}}}}$

Applying this, y' will become

$\displaystyle{y}'=\frac{{\frac{{1}}{{{{\cosh}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}}}{{{2}\cdot\frac{{\sinh{{\left(\frac{{x}}{{2}}\right)}}}}{{\cosh{{\left(\frac{{x}}{{2}}\right)}}}}}}$

$\displaystyle{y}'=\frac{{\frac{{1}}{{{{\cosh}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}}}{{\frac{{{2}{\sinh{{\left(\frac{{x}}{{2}}\right)}}}}}{{\cosh{{\left(\frac{{x}}{{2}}\right)}}}}}}$

$\displaystyle{y}'=\frac{{1}}{{{{\cosh}}^{{2}}{\left(\frac{{x}}{{2}}\right)}}}\cdot\frac{{\cosh{{\left(\frac{{x}}{{2}}\right)}}}}{{{2}{\sinh{{\left(\frac{{x}}{{2}}\right)}}}}}$

$\displaystyle{y}'=\frac{{1}}{{\cosh{{\left(\frac{{x}}{{2}}\right)}}}}\cdot\frac{{1}}{{{2}{\sinh{{\left(\frac{{x}}{{2}}\right)}}}}}$

$\displaystyle{y}'=\frac{{1}}{{{2}{\sinh{{\left(\frac{{x}}{{2}}\right)}}}{\cosh{{\left(\frac{{x}}{{2}}\right)}}}}}$

Then, apply the identity

$\displaystyle{\sinh{{\left({2}{u}\right)}}}={2}{\sinh{{\left({u}\right)}}}{\cos{{\left({u}\right)}}}$

So y' will be

$\displaystyle{y}'=\frac{{1}}{{\sinh{{\left({2}\cdot\frac{{x}}{{2}}\right)}}}}$

$\displaystyle{y}'=\frac{{1}}{{\sinh{{\left({x}\right)}}}}$

Therefore, the derivative of the given function is $\displaystyle{y}'=\frac{{1}}{{\sinh{{\left({x}\right)}}}}$ .

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Saturday, December 11, 2010

$\displaystyle{y}={\ln{{\left({\tanh{{\left(\frac{{x}}{{2}}\right)}}}\right)}}}$ 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?

Saturday, December 11, 2010

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?

$\displaystyle{y}={\ln{{\left({\tanh{{\left(\frac{{x}}{{2}}\right)}}}\right)}}}$ 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?