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Using the definition of the derivative function of a real variable actual value of the...

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juniorsilvamath | Student, Undergraduate | Honors

Posted August 8, 2012 at 8:45 PM via web

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Using the definition of the derivative function of a real variable actual value of the first derivative of function: f (x) = (x-6)/(x +3)

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embizze | High School Teacher | (Level 1) Educator Emeritus

Posted August 8, 2012 at 9:15 PM (Answer #1)

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The formal definition of the derivative of a function is:

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

To find the derivative of `f(x)=(x-6)/(x+3)` :

`f'(x)=lim_(h->0)(((x+h)-6)/((x+h)+3)-((x-6))/((x+3)))/h`

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

** Common denominator for fractios and `(a/b)/h=a/(bh)` **

`=lim_(h->0)(x^2+3x+hx+3h-6x-18-[x^2+hx+3x-6x-6h-18])/(h(x+h+3)(x+3))`

`=lim_(h->0)(9h)/(h(x+h+3)(x+3))`  Now let h=0

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

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Thus `f(x)=(x-6)/(x+3)==>f'(x)=9/((x+3)^2)`

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