Question

Using the limit definition for derivatives, how do you find the derivative of (a) `f(x)=x^2-6x`; (b) `f(x)=4sqrt(x)`?

Answer 1

1) `f'(x) = 2x - 6`

2) `f'(x) = 2/sqrt(x)`

Explanation:

1) `f(x) = x^2 - 6x`

According to the limit definition, the derivative of `f(x)` is

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

To compute `f(x+h)`, plug `x+h` for every occurance of `x` in `f(x)`:

`f(x+h) = (x+h)^2 - 6(x+h)`

Thus, you can compute your derivative as follows:

`f'(x) = lim_(h-> 0) ((x+h)^2 - 6(x+h) - (x^2 - 6x))/h`

`= lim_(h-> 0) (x^2 + 2xh + h^2 - 6x - 6h - x^2 + 6x)/h`

`= lim_(h-> 0) (color(blue)(cancel(x^2)) + 2xh + h^2 - color(red)(cancel(6x)) - 6h - color(blue)(cancel(x^2)) + color(red)(cancel(6x)))/h`

`= lim_(h-> 0) (2xh + h^2 - 6h)/h`

... factor `h` in the numerator...

`= lim_(h-> 0) (h(2x + h - 6))/h`

... cancel `h`...

`= lim_(h-> 0) (2x + h - 6)`

... apply the limit, so plug `h=0`...

`= 2x - 6`

So, we have found that

`f'(x) = 2x -6`.

================================

2) `f(x) = 4 sqrt(x)`

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

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

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

... multiply the numerator and the denominator with `(sqrt(x+h) + sqrt(x))`...

`= lim_(h-> 0) (4(sqrt(x+h) - sqrt(x))color(blue)((sqrt(x+h) + sqrt(x))))/(h color(blue)((sqrt(x+h) + sqrt(x))))`

... use the formula `(a+b)(a-b) = a^2 - b^2` to expand the numerator...

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

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

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

... cancel `h`....

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

... apply the limit, so plug `h=0`...

`= 4 / (sqrt(x + 0) + sqrt(x))`

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

`= 2 / sqrt(x)`

So you have found your derivative and it's

`f'(x) = 2 / sqrt(x)`

Answer 2

If `y=x^2-6x` then `dy/dx= 2x-6`

If `f(x)=4sqrt(x)` then `f'(x)=2/sqrt(x)`

Explanation:

Question 1: `color(black)(y=x^2-6x)`
(using limit definition for derivative)
`dy/dx = lim_(hrarr0)(((x+h)^2-6(x+h)) - (x^2-6x))/h`

`color(white)("XXX")=lim_(hrarro)(cancel(x^2)+2hx+h^2cancel(-6x)-6h-cancel(-x^2)cancel(+6x))/h`

`color(white)("XXX")=lim_(hrarr0)(2cancel(h)x-6cancel(h))/cancel(h)`

`color(white)("XXX")=2x-6`

Question 2: `color(black)(f(x)=4sqrt(x))`
(using limit definition for derivative)
`f'(x)=lim_(hrarr0)=(4sqrt(x+h)-4sqrt(x)/h`

`color(white)("XXX")=4lim_(hrarr0)((sqrt(x+h)-sqrt(x))/h)*((sqrt(x+h)+sqrt(x))/(sqrt(x+h)+sqrt(x)))`

`color(white)("XXX")=4lim_(hrarr0)(x+h-x)/(hsqrt(x+h)+sqrt(x))`

`color(white)("XXX")=4lim_(hrarr0)(cancel(h))/(cancel(h)sqrt(x+h)+sqrt(x))`

`color(white)("XXX")=4*1/(2sqrt(x))`

`color(white)("XXX")=2/sqrt(x)`