Calculus of a Single Variable, Chapter 5, 5.5, Section 5.5, Problem 55

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))

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