Question

Using the limit definition, how do you find the derivative of `f(x) = x^2+3x+1`?

Answer 1

`f^'(x)=2x+3`

Explanation:

`f(x)=x^2+3x+1`
`f^'(x)=Lim_(hrarr0)(f(x+h)-f(x))/h`
`impliesf^'(x)=Lim_(hrarr0){(x+h)^2+3(x+h)+1-(x^2+3x+1)}/h`
`impliesf^'(x)=Lim_(hrarr0){(x+h)^2+3(x+h)+1-x^2-3x-1}/h`
`impliesf^'(x)=Lim_(hrarr0)(x^2+2xh+h^2+3x+3h+1-x^2-3x-1}/h`
`impliesf^'(x)=Lim_(hrarr0)(2xh+h^2+3h}/h`
`impliesf^'(x)=Lim_(hrarr0)(h(2x+h+3)}/h`
`impliesf^'(x)=Lim_(hrarr0)(2x+h+3)`
`impliesf^'(x)=2x+0+3`
`impliesf^'(x)=2x+3`