# `f(x) = ln(sinhx)` Find the derivative of the function

`f(x)=ln(sinh(x))`

Take note that 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

`f'(x)=d/dx[ln(sinh(x))]`

`f'(x)=1/(sinh(x))* d/dx[sinh(x)]`

To take the derivative of hyperbolic sine, apply the formula

• `d/dx[sinh(u)] =cosh(u)*(du)/dx`

So f'(x) will become

`f'(x) =1/(sinh(x))* cosh(x)* d/dx(x)`

`f'(x)=1/(sinh(x))* cosh(x)*1`

`f'(x)= cosh(x)/sinh(x)`

Since...

`f(x)=ln(sinh(x))`

Take note that 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

`f'(x)=d/dx[ln(sinh(x))]`

`f'(x)=1/(sinh(x))* d/dx[sinh(x)]`

To take the derivative of hyperbolic sine, apply the formula

• `d/dx[sinh(u)] =cosh(u)*(du)/dx`

So f'(x) will become

`f'(x) =1/(sinh(x))* cosh(x)* d/dx(x)`

`f'(x)=1/(sinh(x))* cosh(x)*1`

`f'(x)= cosh(x)/sinh(x)`

Since the ratio of hyperbolic cosine to hyperbolic sine is equal to hyperbolic cotangent, the f'(x) will simplify to

`f'(x) =coth(x)`

Therefore, the derivative of the given function is `f'(x)=coth(x)` .

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