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Sholpan [36]
3 years ago
13

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

Mathematics
1 answer:
Sholpan [36]3 years ago
5 0
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}}
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