How do you use the limit definition to find the derivative of #y=sqrt(-4x-4)#?

This is asking us to use the formula #f'(x) = lim_(h->0) (f(x + h) - f(x))/h#.

#f'(x) = lim_(h->0)(sqrt(-4(x + h) -4) - sqrt(-4x - 4))/h#

#f'(x) = lim_(h->0) (sqrt(-4x -4h - 4) - sqrt(-4x - 4))/h#

If we multiply the numerator and the denominator by the conjugate of the numerator, or #sqrt(-4x - 4h - 4) + sqrt(-4x - 4)#, we get:

#f'(x) = lim_(h->0) (sqrt(-4x -4h - 4) - sqrt(-4x - 4))/h ( (sqrt(-4x - 4h - 4) + sqrt(-4x - 4))/(sqrt(-4x - 4h -4) + sqrt(-4x- 4)))#

#f'(x) = lim_(h->0) (-4x - 4h - 4 - (-4x - 4))/(h(sqrt(-4x- 4h - 4) + sqrt(-4x - 4)))#

#f'(x) = lim_(h->0) (-4x - 4h - 4 + 4x + 4)/(h(sqrt(-4x - 4h -4) +sqrt(-4x - 4)))#

#f'(x) = lim_(h->0) (-4h)/(h(sqrt(-4x - 4h - 4) + sqrt(-4x -4))#

#f'(x) = lim_(h->0) -4/(sqrt(-4x - 4h - 4) + sqrt(-4x - 4))#

#f'(x) = -4/(sqrt(-4x - 4(0) - 4) + sqrt(-4x - 4)#

#f'(x) = -4/(2sqrt(-4x - 4))#

#f'(x) = -2/sqrt(-4x - 4)#

If we try getting the derivative using the chain rule, just for verification, we get:

#y' = (d/dx(-4x - 4))/(2sqrt(-4x - 4)#

#y' = -4/(2sqrt(-4x -4))#

#y' = -2/sqrt(-4x- 4)#

As derived above.

Hopefully this helps!