Question

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

Answer 1

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!