# Prove, from first principles, that the derivative of 5xcubed is 15x squared. How would I do this?

### Dan Kozlen

The formula for derivatives from first principles is lim(h to 0) of [f(x+h)-f(x)]/h. So in this case f(x) is 5xcubed, and we want to show that when we use 5xcubed as f(x) in the formula and let h go to 0 we get 15xsquared. Since we're substituting 5xcubed as f(x), we know that 5(x+h)cubed is f(x+h). Expanding 5(x+h)cubed on the top of the fraction gives us: [5(xcubed + 3xsquared*h + 3x*hsquared + hcubed) - 5xcubed]/h. Once you multiply the 5 into the bracket, the 5xcubeds will cancel and then you can cancel an h from each remaining term on the top with the h underneath. This gives: 15xsquared + 15xh + 5hsquared Now as h goes to zero the first term doesn't change since it has no h but the second and third terms also go to zero. Leaving just 15xsquared.

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