Question

How do you find the instantaneous rate of change of `2/(sqrt(x) + 1)` using the `lim_(h->0)` method?

Answer 1

`(-1)/(sqrtx(sqrtx+1)^2)`

See explanation for method.

Explanation:

The limit definition of the derivative states that the derivative of a function `f` is given by:

`f'(x)=lim_(hrarr0)(f(x+h)-f(x))/h`

So here, where `f(x)=2/(sqrtx+1)`, we see that:

`f'(x)=lim_(hrarr0)(2/(sqrt(x+h)+1)-2/(sqrtx+1))/h`

Now, just power through some algebra:

`f'(x)=2lim_(hrarr0)((sqrtx+1-(sqrt(x+h)+1))/((sqrt(x+h)+1)(sqrtx+1)))/h`

`f'(x)=2lim_(hrarr0)(sqrtx-sqrt(x+h))/(h(sqrt(x+h)+1)(sqrtx+1))`

Multiply the numerator and denominator by the conjugate of the numerator:

`f'(x)=2lim_(hrarr0)((sqrtx-sqrt(x+h))(sqrtx+sqrt(x+h)))/(h(sqrt(x+h)+1)(sqrtx+1)(sqrtx+sqrt(x+h)))`

`f'(x)=2lim_(hrarr0)(x-(x+h))/(h(sqrt(x+h)+1)(sqrtx+1)(sqrtx+sqrt(x+h)))`

`f'(x)=2lim_(hrarr0)(-h)/(h(sqrt(x+h)+1)(sqrtx+1)(sqrtx+sqrt(x+h)))`

`f'(x)=2lim_(hrarr0)(-1)/((sqrt(x+h)+1)(sqrtx+1)(sqrtx+sqrt(x+h)))`

Now we can evaluate the limit, since `h` is out of the denominator:

`f'(x)=(-2)/((sqrt(x+0)+1)(sqrtx+1)(sqrtx+sqrt(x+0))`

`f'(x)=(-2)/((sqrtx+1)(sqrtx+1)(sqrtx+sqrtx)`

`f'(x)=(-1)/(sqrtx(sqrtx+1)^2)`