How do you find the derivative of #f(x)=4/(sqrtx-5)# using the limit definition?

#f'(x)=lim_(hrarr0)(4/(sqrt(x+h)-5)-4/(sqrt(x)-5))/h# (difference quotient)

#f'(x)=lim_(hrarr0)((4(sqrt(x)-5))/((sqrt(x+h)-5)(sqrt(x)-5))-(4(sqrt(x+h)-5))/((sqrt(x+h)-5)(sqrt(x)-5)))/h# (combine numerator into 1 fraction)

#f'(x)=lim_(hrarr0)(4sqrt(x)-20-4sqrt(x+h)+20)/((sqrt(x+h)-5)(sqrt(x)-5)h)# (multiply denominator of the fraction in the numerator to the denominator of the larger fraction)

#f'(x)=lim_(hrarr0)(4sqrt(x)-4sqrt(x+h))/((sqrt(x+h)-5)(sqrt(x)-5)h)#

#f'(x)=lim_(hrarr0)((4sqrt(x)-4sqrt(x+h))(4sqrt(x)+4sqrt(x+h)))/((sqrt(x+h)-5)(sqrt(x)-5)h(4sqrt(x)+4sqrt(x+h)))# (conjugate)

#f'(x)=lim_(hrarr0)(16x-16x-16h)/((sqrt(x+h)-5)(sqrt(x)-5)h(4sqrt(x)+4sqrt(x+h)))#

#f'(x)=lim_(hrarr0)(-16h)/((sqrt(x+h)-5)(sqrt(x)-5)h(4sqrt(x)+4sqrt(x+h)))#

#f'(x)=lim_(hrarr0)(-16)/((sqrt(x+h)-5)(sqrt(x)-5)(4sqrt(x)+4sqrt(x+h)))# (remove h from both numerator and denominator)

now you can use direct substitution:
#f'(x)=(-16)/((sqrt(x+0)-5)(sqrt(x)-5)(4sqrt(x)+4sqrt(x+0)))#

#f'(x)=(-16)/((sqrt(x)-5)(sqrt(x)-5)(4sqrt(x)+4sqrt(x)))#

#f'(x)=(-16)/((sqrt(x)-5)(sqrt(x)-5)(8sqrt(x)))#

#f'(x)=(-2)/((sqrt(x)-5)(sqrt(x)-5)(sqrt(x)))#