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.](https://www.tinkutara.com/question/Q127731.png)
$$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} \:……..\left[\mathrm{1}\right] \\ $$$$\mathrm{At}\:\mathrm{Q},\:\mathrm{y}\:+\:\delta\mathrm{y}\:=\:\left(\mathrm{x}+\delta\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{y}+\delta\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}.\delta\mathrm{x}+\left(\delta\mathrm{x}\right)^{\mathrm{2}} …..\left[\mathrm{2}\right] \\ $$$$\mathrm{Subtracting}\:\left[\mathrm{1}\right]\:\mathrm{from}\:\left[\mathrm{2}\right] \\ $$$$\mathrm{y}+\delta\mathrm{y}=\:\mathrm{x}^{\mathrm{2}} +\mathrm{2x}.\delta\mathrm{x}+\left(\delta\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{y}\:\:\:\:\:\:\:\:\:=\:\mathrm{x}^{\mathrm{2}} \\ $$$$−−−−−−−−−−−−−−− \\ $$$$\delta\mathrm{y}\:=\:\mathrm{2x}.\delta\mathrm{x}+\left(\delta\mathrm{x}\right)^{\mathrm{2}} \\ $$$$\mathrm{Divide}\:\mathrm{through}\:\mathrm{by}\:\delta\mathrm{x} \\ $$$$\frac{\delta\mathrm{y}}{\delta\mathrm{x}}\:=\:\mathrm{2x}\:+\delta\mathrm{x} \\ $$$$\mathrm{If}\:\delta\mathrm{x}\rightarrow\mathrm{0},\frac{\delta\mathrm{y}}{\delta\mathrm{x}}\rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:\mathrm{and}\:\therefore\:\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{2x} \\ $$$$\therefore\:\mathrm{If}\:\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} ,\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{2x}. \\ $$$$ \\ $$