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


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

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




Now


`\displaystyle(\mathrm dy)/(\mathrm dx)=\lim_(h\to0)((x^2+2xh+h^2+1)-(x^2+1))/(h\left(√(x^2+2xh+h^2+1)+√(x^2+1)\right))`

`=\displaystyle\lim_(h\to0)(2xh+h^2)/(h\left(√(x^2+2xh+h^2+1)+√(x^2+1)\right))`

`=\displaystyle\lim_(h\to0)(2x+h)/(√(x^2+2xh+h^2+1)+√(x^2+1))`

As
`h\to0`, in the numerator we have
`2x+h\to2x`; in the denominator we have
`√(x^2+2xh+h^2+1)\to√(x^2+1)`. So the limit is


`(2x)/(2√(x^2+1))=\frac x{√(x^2+1)}`