# `f(x) = x^2 - 5, c = 3` Use the alternate form of the derivative to find the derivative at x = c (if it exists)

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the alternate method to find the derivative of the function is the limit form.

**By limit process, the derivative of a function f(x) is :-**

**f'(x) = lim h --> 0 [{f(x+h) - f(x)}/h]**

Now, the given function is :-

f(x) = (x^2) - 5

thus, f'(x) = lim h --> 0 [{(x+h)^2 - 5 - {(x^2) - 5}}/h]

or, f'x) = lim h --> 0 [{(h^2) + 2hx}/h]

or, f'(x) = lim h --> 0 [{2x + h}]

putting the value of h = 0 in the above expression we get

f'(x) = 2x

Thus, f'(c) = f'(3) = 2*3 = 6

Using the alternative definition of the derivative, the approximation of the derivative given by

`lim_(x->3)` `(f(x) -f(c))/(x-c)` = `lim_(x->3)(x^2 -5 - (3^2 -5))/(x-3)` ` `

`=(x^2 -5 -4)/(x-3) `

`=(x^2 - 9)/(x-3) `

`=((x -3)(x+3))/(x-3) `

`=(x+3) `

` ` `=(3+3) `

`=6 `

`lim_(x->2)` `(f(x) -f(c))/(x-c)`

Using the alternative definition of the derivative, the approximation of the derivative given by

= `(3 + 3)`

= `6`