`g(x) = x^2 - x, c = 1` Use the alternate form of the derivative to find the derivative at x = c (if it exists)

Textbook Question

Chapter 2, 2.1 - Problem 66 - Calculus of a Single Variable (10th Edition, Ron Larson).
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leonard-chen | (Level 2) Adjunct Educator

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`lim_(x->1)(f(x)-f(c))/(x-c)`

`lim_(x->1)(x^2 - x)/(x - 1)`

``Simplify:

`lim_(x->1)(x(x-1))/(x-1)`

`lim_(x->1)(x) = 1`

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hkj1385 | (Level 1) Assistant Educator

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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) - x

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

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

or, `f'(x) = lim_(h -> 0) [{2x + h - 1}]`

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

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

Thus, `f'(c) = f'(1) = 2*1 - 1 = 1`

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