Question

Ho do I use the limit definition of derivative to find `f'(x)` for `f(x)=3x^2+x` ?

Answer 1

By Power Rule, we know that we are supposed to get `f'(x)=6x+1`.

Let us find it using the definition
`f'(x)=lim_{h to 0}{f(x+h)-f(x)}/{h}`

Let us first find the difference quotient
`{f(x+h)-f(x)}/{h}={3(x+h)^2+(x+h)-[3x^2+x]}/{h}`
By simplifying the numerator,
`={3(x^2+2xh+h^2)+x+h-3x-x}/{h}`
`={3x^2+6xh+3h^2+h-3x^2}/{h}`
`={6xh+3h^2+h}/{h}`
By factoring out `h` from the numerator,
`={h(6x+3h+1)}/h`
By cancelling out `h`'s,
`=6x+3h+1`

Hence,
`f'(x)=lim_{h to 0}(6x+3h+1)=6x+3(0)+1=6x+1`