Question

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

Answer 1

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

Explanation:

`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)))`