I also use #d/dx(sqrtx) = 1/(2sqrtx)#

so, the chain rule gives us: #d/dx(sqrtu) = 1/(2sqrtu) (du)/dx#

#f(x) = (sqrt(x+13))/((x-4)sqrt(2x+1))#

For #f(x) = T/B#, the quotient rule gives us: #f'(x) = (T'B-TB')/(B^2)#

To find the derivative of the denominator, we'll use the product rule:

for #g(x) = FS#, we have #g'(x) = F'S+FS'#

So, we get:

#f'(x) = (1/(2sqrt(x+13))((x-4)sqrt(2x+1))-sqrt(x+13) [ (1)sqrt(2x+1)+(x-4)1/(2sqrt(2x+1))*2 ])/((x-4)sqrt(2x+1))^2#

That's it for the calculus. Now we need to do some algebra to write this in a nicer (or at least, less messy) form.

#([1/(2sqrt(x+13))((x-4)sqrt(2x+1))]-[sqrt(x+13)sqrt(2x+1)] - [(sqrt(x+13)(x-4))/(sqrt(2x+1))] )/((x-4)^2(2x+1))#

We'll clear the fractions in the numerator by multiplying the numerator and the denominator by:

#(2sqrt(x+13)sqrt(2x+1))#

We get:

#((x-4)(2x+1)-2(x+13)(2x+1)-2(x+13)(x-4) )/(2sqrt(x+13)sqrt(2x+1)(x-4)^2(2x+1))#

# = ((2x^2-7x-4)-2(2x^2+27x+13)-2(x^2+9x-52))/(2sqrt(x+13)sqrt(2x+1)(x-4)^2(2x+1))#

# = (-4x^2 - 79x +74)/(2sqrt(x+13)sqrt(2x+1)(x-4)^2(2x+1))#

(I think that's correct)