Question

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

Answer 1

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

Explanation:

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`