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

Textbook Question

Chapter 2, 2.1 - Problem 65 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

2 Answers | Add Yours

dh246's profile pic

dh246 | (Level 1) Adjunct Educator

Posted on

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`

hkj1385's profile pic

hkj1385 | (Level 1) Assistant Educator

Posted on

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

We’ve answered 318,911 questions. We can answer yours, too.

Ask a question