Question

How do you find f'(x) using the definition of a derivative `f(x)=6 x + 2 /sqrt{x}`?

Answer 1

Determination, attention to detail and algebra. See below.

Explanation:

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

`= lim_(hrarr0) ([6(x+h)+2/sqrt(x+h)] - [ 6x+2/sqrtx])/h`

This limit looks hard to evaluate, but we can split it into two limits:

`= lim_(hrarr0)(6(x+h)-6x)/h + lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h`

The first limit evaluates to `6`, so let's focus on the second limit.

`lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h`

This is a combination of problems you've probably seen before.

Like `lim_(hrarr0)(1/(x+h)-1/x)/h` and `lim_(hrarr0)(sqrt(x+h)-sqrtx)/h`.

We should try the methods that worked for those problems.

First, let's get a single rational expression in the numerator.

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

And now a single rational expression.

`= lim_(hrarr0)((2sqrtx-2sqrt(x+h))/(hsqrt(x+h)sqrtx))`

If we try to evaluate the limit, we still get the indeterminate form `0/0`.

I suppose we could try rationalizing the numerator to see if that helps.

`= lim_(hrarr0)((2sqrtx-2sqrt(x+h)))/(hsqrt(x+h)sqrtx) * ((2sqrtx+2sqrt(x+h)))/((2sqrtx+2sqrt(x+h)))`

`= lim_(hrarr0)((4x-4(x+h)))/(hsqrt(x+h)sqrtx (2sqrtx+2sqrt(x+h))`

`= lim_(hrarr0)(-4cancel(h))/(cancel(h)sqrt(x+h)sqrtx (2sqrtx+2sqrt(x+h))`

The numerator no longer goes to `0`!

That means we can evaluate the limit. (Then we'll add the `6` we got above to finish finding the derivative of `f`).

`lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h = lim_(hrarr0)(-4)/(sqrt(x+h)sqrtx (2sqrtx+2sqrt(x+h))`

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

`= (-4)/ (sqrtxsqrtx (4sqrtx)) = (-1)/(sqrtx)^3`

Finally we finish

`f'(x) = lim_(hrarr0) ([6(x+h)+2/sqrt(x+h)] - [ 6x+2/sqrtx])/h`

`= lim_(hrarr0)(6(x+h)-6x)/h + lim_(hrarr0)(2/sqrt(x+h)-2/sqrtx)/h`

`= 6 - 1/(sqrtx)^3`