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# Can't see why I'm getting the wrong answer (Devirates h definition)

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f'(-4) if f(x) = x^2-x

So I'm calculating

f´(-4) = lim(h goes against 0) (f(-4+h)-f(-4))/h

f´(-4) = lim(h goes against 0) ((-4+h)^2+4+4^2-4)/h

f´(-4) = lim(h goes against 0)(4^2-4^2-8h+h^2)/h

f´(-4) = lim(h goes against 0)h(-8+h)/h

f´(-4) = lim(h goes against 0) -8+h

f´(-4) = -8

I know the answer should be -9 since the diverate of f(x) = x^2-x should be 2x - 1 (-4*2-1) but I can't make it work :(

May 29, 2017

### 2+0 Answers

#1
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By definition, we have

[ (x + h)^2  - (x + h)  -  ( x^2 - x )  ]  /  h   =

[ x^2 + 2xh + h^2 - x - h  -  x^2  + x ]  /  h  =

[ 2xh  + h^2 -  h]  / h   =

[ (h)( 2x  + h  -  1 )]  / h   =     divide by h on top/bottom

[ 2x  + h  -  1 ]        let  h →  0     and we have

[ 2x  -  1 ]  =   f ' (x)

So

f '(-4)   =    2(-4)  -  1    =  - 8  -  1    =    - 9   May 29, 2017
#2
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Riight, forgot that you add (x+h) into all x terms in the function, thank you!

May 29, 2017