# What is derivative f(x)=ln(sin(x-1))+ln(sin(x+1))?

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### 1 Answer

You need to notice that the function `f(x)` consists of composed functions, hence, you need to use the chain rule to evaluate its derivative, such that:

`f'(x) = 1/(sin(x-1))*(sin(x-1))'*(x-1)' + 1/(sin(x+1))*(sin(x+1))'*(x+1)' `

`f'(x) = 1/(sin(x-1))*(cos(x-1))*1+ 1/(sin(x+1))*(cos(x+1))*1`

`f'(x) = (cos(x-1))/(sin(x-1)) + (cos(x+1))/(sin(x+1))`

`f'(x) = (cos(x-1)sin(x+1) + cos(x+1)sin(x-1))/(sin(x-1)sin(x+1))`

Using the trigonometric identity `sin a*cos b + sin b*cos a = sin(a+b)` yields:

`f'(x) = (sin(x + 1+ x - 1))/(sin(x-1)sin(x+1))`

Reducing duplicate members yields:

`f'(x) = (sin 2x)/(sin(x-1)sin(x+1))`

You may also use the trigonometric identity `cot x = cos x/sin x` , from the point `f'(x) = (cos(x-1))/(sin(x-1)) + (cos(x+1))/(sin(x+1))` , such that:

`f'(x) = cot(x-1) + cot(x+1)`

**Hence, evaluating the derivative of the given function yields `f'(x) = cot(x-1) + cot(x+1)` that can be converted into the alternative answer **`f'(x) = (sin 2x)/(sin(x-1)sin(x+1)).`

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