How do we find the differential of $y=x^2+1$ from first principle?

Question

8083 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-19T06:48:53

$(dy)/(dx)=2x-2$

According to fist principle, if $y=f(x)$

then $(dy)/(dx)=Lt_(h->0)(f(x+h)-f(x))/h$

Here $f(x)=(x-1)^2+1$

therefore $f(x+h)=(x+h-1)^2+1$

and $f(x+h)-f(x)$

= $(x+h-1)^2+1-(x-1)^2+1$

= $(x+h-1)^2-(x-1)^2$

= $(x+h-1+x-1)(x+h-1-x+1)$
- using $a^2-b^2=(a+b)(a-b)$

= $h(2x+h-2)$

and $(dy)/(dx)=Lt_(h->0)(h(2x+h-2))/h$

= $Lt_(h->0)(2x+h-2)$

= $2x-2$

How do we find the differential of y=x^2+1y=x^2+1 from first principle?