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

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lemjay eNotes educator| Certified Educator

`y=ln(tanh(x/2))`

The derivative formula of natural logarithm is

  • `d/dx[ln(u)] = 1/u*(du)/dx`

Applying this formula, the derivative of the function will be

`y' = d/dx [ln(tanh(x/2))]`

`y' = 1/(tanh(x/2)) * d/dx[tanh(x/2)]`

To take the derivative of hyperbolic tangent, apply the formula

  • `d/dx[tanh(u)] = sec h^2 (u) * (du)/dx`

So y' will become

`y'= 1/(tanh(x/2)) * sec h^2 (x/2) * d/dx(x/2)`

`y' = 1/(tanh(x/2)) *sec h^2(x/2) * 1/2`

`y'=(sec h^2(x/2))/(2tanh(x/2))`

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

  • `sec h(u) = 1/cosh(u)`
  • `tanh(u)=sinh(u)/cosh(u)`

Applying this, y' will become

`y'= (1/(cosh^2(x/2)))/(2*sinh(x/2)/cosh(x/2))`

`y'= (1/(cosh^2(x/2)))/((2sinh(x/2))/cosh(x/2))`

`y'=1/(cosh^2(x/2)) * cosh(x/2)/(2sinh(x/2))`

`y'=1/cosh(x/2) * 1/(2sinh(x/2))`

`y'=1/(2sinh(x/2)cosh(x/2))`

Then, apply the identity

  • `sinh(2u) = 2sinh(u)cos(u)`

So y' will be

`y' = 1/sinh(2*x/2)`

`y'=1/sinh(x)`

 

Therefore, the derivative of the given function is `y'=1/sinh(x)` .

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