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using limits show that derivative of x^2 is 2x.

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PaulRosandro | (Level 1) eNoter

Posted August 30, 2013 at 9:39 AM via web

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using limits show that derivative of x^2 is 2x.

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted August 30, 2013 at 9:45 AM (Answer #1)

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Let us say;

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

Then for a small change `deltax ` in x if we have `deltay` change in y;

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

(2)-(1)

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

`deltay = (x+deltax+x)(x+deltax-x)`

`deltay = (2x+deltax)(deltax)`

The derivative of y is defined as;

`(dy)/(dx) = lim_(deltaxrarr0) (deltay)/(deltax)`

`(dy)/(dx) = lim_(deltaxrarr0) ((2x+deltax)(deltax))/(deltax)`

`(dy)/(dx) = lim_(deltaxrarr0) ((2x+deltax)`

`(dy)/(dx) = 2x+0`

`(dy)/(dx) = 2x`

So the derivative of x^2 is 2x as proved from first principles.

Sources:

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tonys538 | Student, Undergraduate | TA | (Level 1) Valedictorian

Posted February 13, 2015 at 1:53 PM (Answer #2)

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The derivative of a function y = f(x) is given by the limit:

`lim_(h-> 0) (f(x + h)-f(x))/h` ...(1)

Here, the function `f(x) = x^2` .

Substituting this in (1) gives:

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

Use the rule `x^2 - a^2 = (x - a)(x + a)`

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

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

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

Now substitute h = 0

= x + x

= 2x

This shows that the derivative of f(x) = x^2 is f'(x) = 2x

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