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Question Number 127731 by Tosin Okunowo last updated on 01/Jan/21

y = x^2  ........[1]  At Q, y + δy = (x+δx)^2   y+δy = x^2 +2x.δx+(δx)^2 .....[2]  Subtracting [1] from [2]  y+δy= x^2 +2x.δx+(δx)^2   y         = x^2   −−−−−−−−−−−−−−−  δy = 2x.δx+(δx)^2   Divide through by δx  ((δy)/(δx)) = 2x +δx  If δx→0,((δy)/(δx))→(dy/dx) and ∴ (dy/dx)=2x  ∴ If y = x^2 ,(dy/dx) = 2x.

$$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:........\left[\mathrm{1}\right] \\ $$$$\mathrm{At}\:\mathrm{Q},\:\mathrm{y}\:+\:\delta\mathrm{y}\:=\:\left(\mathrm{x}+\delta\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{y}+\delta\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}.\delta\mathrm{x}+\left(\delta\mathrm{x}\right)^{\mathrm{2}} .....\left[\mathrm{2}\right] \\ $$$$\mathrm{Subtracting}\:\left[\mathrm{1}\right]\:\mathrm{from}\:\left[\mathrm{2}\right] \\ $$$$\mathrm{y}+\delta\mathrm{y}=\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}.\delta\mathrm{x}+\left(\delta\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{y}\:\:\:\:\:\:\:\:\:=\:\mathrm{x}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−− \\ $$$$\delta\mathrm{y}\:=\:\mathrm{2x}.\delta\mathrm{x}+\left(\delta\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{Divide}\:\mathrm{through}\:\mathrm{by}\:\delta\mathrm{x} \\ $$$$\frac{\delta\mathrm{y}}{\delta\mathrm{x}}\:=\:\mathrm{2x}\:+\delta\mathrm{x} \\ $$$$\mathrm{If}\:\delta\mathrm{x}\rightarrow\mathrm{0},\frac{\delta\mathrm{y}}{\delta\mathrm{x}}\rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:\mathrm{and}\:\therefore\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{2x} \\ $$$$\therefore\:\mathrm{If}\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} ,\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{2x}. \\ $$$$ \\ $$

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