G technical mathematics with calculus volume 10 find the derivative of the function y = sqrt(x^2+1) using limits definition

By definition of the derivative,

[tex]\displaystyle\frac{\mathrm dy}{\mathrm dx}=\lim_{h\to0}\frac{y(x+h)-y(x)}h[/tex]

[tex]\displaystyle\frac{\mathrm dy}{\mathrm dx}=\lim_{h\to0}\frac{\sqrt{(x+h)^2+1}-\sqrt{x^2+1}}h=\lim_{h\to0}\frac{\sqrt{x^2+2xh+h^2+1}-\sqrt{x^2+1}}h[/tex]

Multiply the numerator and denominator by the conjugate of the numerator:

[tex]\dfrac{\sqrt{x^2+2xh+h^2+1}-\sqrt{x^2+1}}h\cdot\dfrac{\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}}{\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}}=\dfrac{(x^2+2xh+h^2+1)-(x^2+1)}{h\left(\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}\right)}[/tex]

Now

[tex]\displaystyle\frac{\mathrm dy}{\mathrm dx}=\lim_{h\to0}\frac{(x^2+2xh+h^2+1)-(x^2+1)}{h\left(\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}\right)}[/tex]
[tex]=\displaystyle\lim_{h\to0}\frac{2xh+h^2}{h\left(\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}\right)}[/tex]
[tex]=\displaystyle\lim_{h\to0}\frac{2x+h}{\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}}[/tex]

As [tex]h\to0[/tex], in the numerator we have [tex]2x+h\to2x[/tex]; in the denominator we have [tex]\sqrt{x^2+2xh+h^2+1}\to\sqrt{x^2+1}[/tex]. So the limit is

[tex]\dfrac{2x}{2\sqrt{x^2+1}}=\dfrac x{\sqrt{x^2+1}}[/tex]