`h(x) = log_3(xsqrt(x-1)/2)` Find the derivative of the function

 The derivative of a function h with respect to x is denoted by `h'(x)` .

 To solve for `h'(x)` for given function `h(x) =log_3(x*sqrt(x-1)/2)` , we apply the derivative for logarithmic functions: `d/(dx) log_a(u)= ((du)/(dx))/(u*ln(a))` .

We may let `u =x*sqrt(x-1)/2` and `a = 3` .

In solving for the derivative of u: `(du)/(dx)` , we apply basic property: `d/(dx) (c* f(x)) = c * d/(dx) f(x)` .

`d/(dx) u = d/(dx)(x*sqrt(x-1)/2)`

`u' = (1/2) * d/(dx)(x*sqrt(x-1))`

Applying the Product Rule for derivative: `d/(dx)(f*g) = f'*g + f*g'` .

Let:

`f = x` then `f'= 1`

`g=sqrt(x-1)` then` g'=1/(2sqrt(x-1))`

Then following the formula: `d/(dx)(f*g) = f'*g + f*g'` , we set-up: 

`u' = (1/2) * d/(dx)(x*sqrt(x-1))`

`u' = (1/2) * [ 1 *(sqrt(x-1)+ (x)*(1/(2sqrt(x-1)))]`

Simplify:

`u'=(1/2) * [ 1 *(sqrt(x-1) *(2sqrt(x-1))/(2sqrt(x-1)) + x/(2sqrt(x-1))]`

    `=(1/2) * [(2(sqrt(x-1))^2)/(2sqrt(x-1)) + x/(2sqrt(x-1))]`

    `=(1/2) * [(2(x-1))/(2sqrt(x-1)) + x/(2sqrt(x-1))]`

   ` =(1/2) * [(2x-2)/(2sqrt(x-1)) + x/(2sqrt(x-1))]`

    `=(1/2) * [(2x-2+x)/(2sqrt(x-1))]`

    `=(1/2) * [(3x-2)/(2sqrt(x-1))]`

  `= (3x-2)/(4sqrt(x-1))`

Applying `u = x*sqrt(x-1)/2` , `a=3` , and `(du)/(dx) or u' =(3x-2)/(4sqrt(x-1))`

with the derivative formula: `d/(dx) log_a(u)= ((du)/(dx))/(u*ln(a))` , we get:

`d/(dx) (log_3(x*sqrt(x-1)/2)) =((3x-2)/(4sqrt(x-1)))/(xsqrt(x-1)/2ln(3))`

This is also the same as:

`h'(x) =(3x-2)/(4sqrt(x-1)) * 1/((xsqrt(x-1)ln(3))/2)`

Flip the `xsqrt(x-1)ln(3))/2` to proceed to multiplication:

`h'(x) =(3x-2)/(4sqrt(x-1)) * ( 1* 2/(xsqrt(x-1)/ln(3)))`

Multiply across:

`h'(x) = (2(3x-2))/(4x(sqrt(x-1))^2ln(3))`

Simplify by applying: `2/4 =1/2 ` and `(sqrt(x-1))^2 = x-1` .

FINAL ANSWER: 

`h'(x)=(3x-2)/(2x(x-1)ln(3))`