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)))
