The derivative of the quotient is determined by using the definition of derivatives:
color(red)(f'(x)=lim_(h->0)(f(x+h)-f(x))/h)
In this exercise we will also apply this polynomial identity to expand:
color(brown)((a^2-b^2)=(a-b)(a+b))
f'(x)=lim_(h->0)(4/(sqrt(x+h))-4/sqrtx)/h
f'(x)=lim_(h->0)((4sqrtx-4sqrt(x+h))/(sqrt(x+h)sqrtx))/h
f'(x)=lim_(h->0)((4sqrtx-4sqrt(x+h)))/(h(sqrt(x+h)sqrtx)
To get rid of the square root in the numerator we will multiply by its conjugate color(blue)(4sqrtx+4sqrt(x+h))
f'(x)=lim_(h->0)((4sqrtx-4sqrt(x+h))(color(blue)(4sqrtx+4sqrt(x+h))))/(h(sqrt(x+h)sqrtx)(color(blue)(4sqrtx+4sqrt(x+h))))
f'(x)=lim_(h->0)((4sqrtx)^2-(4sqrt(x+h))^2)/(h(sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))
f'(x)=lim_(h->0)(16x-16(x+h))/(h(sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))
f'(x)=lim_(h->0)(cancel(16x)cancel(-16x)-16h)/(h(sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))
f'(x)=lim_(h->0)(-16cancel(h))/(cancelh(sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))
Now let us find the limit color(green)(lim_(h->0))
f'(x)=lim_(h->0)(-16)/((sqrt(x+h)sqrtx)(4sqrtx+4sqrt(x+h)))
f'(x)=(-16)/((sqrt(x+0)sqrtx)(4sqrtx+4sqrt(x+0)))
f'(x)=(-16)/((sqrt(x)sqrtx)(4sqrtx+4sqrt(x)))
f'(x)=(-cancel16)/(x(cancel8sqrtx))
f'(x)=(-2)/(xsqrtx)
