Differentiate by definition

[calcnovis]

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

 

[DrPhil]

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
No, should be:x² + 2x(Δx) + (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

You had extra terms when you squared the binomial. Fixing that should get rid of the square in your answer.

 

[JeffM]

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 NO \(\displaystyle (x + \Delta x)^2 - 2(x + \Delta x ) + 2 = x^2 + 2x \Delta x + (\Delta x)^2 -2x - 2\Delta x + 2.\)
Δy=2x²Δx+2x(Δx)²+(Δx)²-2Δx SO \(\displaystyle \Delta y = x^2 + 2x \Delta x + (\Delta x)^2 -2x - 2\Delta x + 2 - (x^2 - 2x + 2) = 2x \Delta x + (\Delta x)^2 -2\Delta x.\)

Δy/Δx=2x²+2xΔx+(Δx)²-2 SO \(\displaystyle \dfrac{\Delta y}{\Delta x} = \dfrac{2x \Delta x + (\Delta x)^2 - 2\Delta x}{\Delta x} = 2x + \Delta x - 2.\)
dy/dx=lim/Δx->0 [2x²+2xΔx+(Δx)²-2] SO \(\displaystyle \dfrac{dy}{dx} =\displaystyle \lim_{\Delta x \rightarrow 0}(2x + \Delta x - 2) = \left(\lim_{\Delta x \rightarrow 0}2x\right) + \left(\lim_{\Delta x \rightarrow 0}\Delta x\right) + \left(\lim_{\Delta x \rightarrow 0}2\right).\)
dy/dx=2x²-2 SO \(\displaystyle \dfrac{dy}{dx} = 2x + 0 - 2 = 2x - 2.\)

You must be careful with your algebra in calculus.