Question

How do you use the limit definition to find the derivative of `y=sqrt(-4x-2)`?

Answer 1

See below.

Explanation:

The limit definition of the derivative is given by:

`f(x)=lim_(h->0)(f(x+h)-f(x))/h`

We have `f(x)=y=sqrt(-4x-2)`

Putting this into the above definition:

`f(x)=lim_(h->0)(sqrt(-4(x+h)-2)-sqrt(-4x-2))/h`

Now we attempt to simplify.

`=>lim_(h->0)(sqrt(-4x-4h-2)-sqrt(-4x-2))/h`

Clearly we must get `h` out of the denominator. We can do this by multiplying the numerator and denominator by the conjugate of the numerator.

`=>lim_(h->0)[((sqrt(-4x-4h-2)-sqrt(-4x-2))/h)*(sqrt(-4x-4h-2)+sqrt(-4x-2))/(sqrt(-4x-4h-2)+sqrt(-4x-2)))]`

`=>lim_(h->0)[((-4x-4h-2)-(-4x-2))/(h(sqrt(-4x-4h-2)+sqrt(-4x-2)))]`

`=>lim_(h->0)[-(4cancelh)/(cancelh(sqrt(-4x-4h-2)+sqrt(-4x-2)))]`

`=>lim_(h->0)[-(4)/(sqrt(-4x-4h-2)+sqrt(-4x-2))]`

`=>-(4)/(sqrt(-4x-4(0)-2)+sqrt(-4x-2))`

`=>-(4)/(sqrt(-4x-2)+sqrt(-4x-2))`

`=>-(4)/(2sqrt(-4x-2))`

`=>-(2)/(sqrt(-4x-2))`

We can verify this answer by taking the derivative directly (using the chain rule):

`y=sqrt(-4x-2)`

`=(-4x-2)^(1/2)`

`y'=1/2(-4x-2)^(-1/2)*(-4)`

`=-2/(sqrt(-4x-2))`