Question

How do you use the limit definition of the derivative to find the derivative of `f(x)=x^2+3x`?

Answer 1

`dy/dx=2x+3`

Explanation:

By definition:
`dy/dx=lim_(h->0)(f(x+h)-f(x))/h`

so, with `y=x^2+3x` we have:

`dy/dx=lim_(h->0)(( ((x+h)^2+3(x+h)) -(x^2+3x))/h)`
`dy/dx=lim_(h->0)(( (x^2+2hx+h^2+3x+3h) -(x^2+3x))/h)`
`dy/dx=lim_(h->0)(( x^2+2hx+h^2+3x+3h -x^2-3x)/h)`
`dy/dx=lim_(h->0)(( color(red)cancel(x^2)+2hx+h^2+color(blue)cancel(3x)+3h color(red)cancel(-x^2)color(blue)cancel(-3x))/h)`

`dy/dx=lim_(h->0)(( 2hx+h^2+3h )/h)`
`dy/dx=lim_(h->0)( 2x+h+3 )`
`dy/dx=2x+3`