Cal...
(x+Δx)^2=Χ^2+2(Χ)(ΔΧ)+(ΔΧ)^2 ΑΝD -2(Χ+ΔΧ) = -2Χ-2ΔΧ
CORRECT YOUR SECOND LINE (EXPANSION)
Anyway the final result is ok.
Minoas
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^{2}Δx+2x(Δx)^{2}+(Δx)^{2}-2Δx
Δy/Δx=2x^{2}+2xΔx+(Δx)^{2}-2
dy/dx=lim/Δx->0 [2x^{2}+2xΔx+(Δx)^{2}-2]
dy/dx=2x^{2}-2
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