Question

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

Answer 1

By definition of the derivative,

$\displaystyle\frac{\mathrm dy}{\mathrm dx}=\lim_{h\to0}\frac{y(x+h)-y(x)}h$

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

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

$\dfrac{\sqrt{x^2+2xh+h^2+1}-\sqrt{x^2+1}}h\cdot\dfrac{\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}}{\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}}=\dfrac{(x^2+2xh+h^2+1)-(x^2+1)}{h\left(\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}\right)}$

Now

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

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

$\dfrac{2x}{2\sqrt{x^2+1}}=\dfrac x{\sqrt{x^2+1}}$