Determine f ′(x) from first principles if f (x) = −3x^2 .

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To use first principles, use the formula:

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

h approaches zero as it gets infinitely smaller but never actually reaches zero.

Now whereever you see x in your original equation you will substitute (x+h) instead:

`therefore f(x) = -3x^2` becomes

`f'(x) = lim_(h->0) (-3(x+h)^2 - (-3x^2))/h`

Note how because x was squared we now apply that to (x+h)

Now simplify:

`f'(x) = lim_(h->0) (-3(x^2 +2hx + h^2) + 3x^2)/h`

`therefore = lim_(h->0) (-3x^2-6hx -3h^2 +3x^2)/h`

Note that the `3x^2`  cancel each other out leaving

`= lim_(h->0) (-6hx - 3h^2)/h`

Now factorize using the h only:

`therefore = lim_(h->0) (h(-6x -3h))/h`

Now the h cancels out leaving:

`therefore = lim_(h->0) -6x-3h`

Now as h is approaching zero we substitute that value for h:

`therefore = lim_(h->0) -6x - 3(0)`

`therefore f'(x) = -6x`

Remember to use the `lim_(h->0)` on each line until you get to the end.

Ans: f'(x) = -6x

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`f(x) = y = -3x^2` -----(1)

For a small change of `deltax ` in x let us say change in y is `deltay` .

`y+deltay = -3(x+deltax)^2 ` -----(2)

(2)-(1)

`deltay = -3((x+deltax)^2-x^2)`

`deltay = -3(x^2+2xdeltax+(deltax)^2-x^2)`

`deltay = -3(2xdeltax-(deltax)^2)`

Since `deltax` is a very small value `(deltax)^2~=0.`

`deltay = -3(2xdeltax-(deltax)^2)`

`deltay = -3(2xdeltax)`

`deltay = -6xdeltax`

`(deltay)/(deltax) = -6x`

The derivative of y is defined as;

`y' = lim_(deltaxrarr0)(deltay)/(deltax)`

`y' = lim_(deltaxrarr0)(-6x)`

`y' = -6x`

`y' = f'(x) = -6x`

So the finding of derivative from first principal is complete.

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