1. ## Differentiate by definition

Can someone please confirm I am doing this correctly? Thanks!
Y=x2-2x+2 at P(x,y), let Q (x+Δx, y+Δy) be another point on the curve. Differentiate by definition
y+
Δy=(x+Δx)2 -2(x+Δx)+2
=X2 +2x2 (
Δx)+2x(Δx)2+(Δx)2-2x-2Δx+2
Δy=2x2Δx+2x(Δx)2+(Δx)2-2Δx
Δy/Δx=2x2+2xΔx+(Δx)2-2
dy/dx=lim/
Δx->0 [2x2+2xΔx+(Δx)2-2]
dy/dx=2x2-2

2. ## Re: Differentiate by definition

Cal...

(x+Δx)^2=Χ^2+2(Χ)(ΔΧ)+(ΔΧ)^2 ΑΝD -2(Χ+ΔΧ) = -2Χ-2ΔΧ
Anyway the final result is ok.

Minoas

3. ## Re: Differentiate by definition

Can someone please confirm I am doing this correctly? Thanks!
Y=x^2-2x+2 at P(x,y), let Q (x+Δx, y+Δy) be another point on the curve. Differentiate by definition
y+Δy=(x+Δx)^2 -2(x+Δx)+2
=X^2 +2x^2 (Δx)+2x(Δx)^2+(Δx)^2-2x-2Δx+2
Δy=2x^2Δx+2x(Δx)^2+(Δx)^2-2Δx
Δy/Δx=2x^2+2xΔx+(Δx)^2-2
dy/dx=lim/Δx->0 [2x2+2xΔx+(Δx)2-2]
dy/dx=2x2-2
$y = x^2 - 2x + 2$

$y + \Delta y = (x + \Delta x)^2 - 2(x + \Delta x) +2 = x^2 + 2x\Delta x + (\Delta x)^2 - 2x - 2\Delta x + 2$

$\Delta y = 2x\Delta x + (\Delta x)^2 - 2\Delta x$

$\frac{\Delta y}{\Delta x} = 2x + \Delta x - 2$

$\frac{dy}{dx} = \lim_{\Delta x \rightarrow 0}(2x + \Delta x - 2) = 2x - 2$

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5. ## Re: Differentiate by definition

Originally Posted by MINOANMAN
Cal...

(x+Δx)^2=Χ^2+2(Χ)(ΔΧ)+(ΔΧ)^2 ΑΝD -2(Χ+ΔΧ) = -2Χ-2ΔΧ
Anyway the final result is ok.
No, the final result is not correct.

Minoas

6. ## Re: Differentiate by definition

Originally Posted by HallsofIvy
No, the final result is not correct.
So, then is Jpritch22 correct?

7. ## Re: Differentiate by definition

Halls of Ivy was responding to minoanman's response. My solution should be correct

8. ## Re: Differentiate by definition

Ok great. Thanks for your help!