Question

Find the derivative of `t^3+t^2` using first principles?

Answer 1

`f'(t) = 3t^2+2t`

Explanation:

The definition of the derivative of `y=f(x)` is

`f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h`

So with `f(t) = t^3+t^2` then;

`f(t+h) = (t+h)^3 + (t+h)^2`
`" " = t^3+3y^2h+3th^2+h^3 + (t^2+2ht+h^2)`
`" " = t^3+3t^2h+3th^2+h^3 + t^2+2th+h^2`

And so the derivative of `y=f(x)` is given by:

`f'(t) = lim_(h rarr 0) ( (t^3+3t^2h+3th^2+h^3 + t^2+2th+h^2) - (t^3+t^2) ) / h`

`" " = lim_(h rarr 0) ( t^3+3t^2h+3th^2+h^3 + t^2+2th+h^2 -t^3-t^2 ) / h`

`" " = lim_(h rarr 0) ( 3t^2h+3th^2+h^3 +2th+h^2 ) / h`

`" " = lim_(h rarr 0) ( 3t^2+3th+h^2 +2t+h)`

`" " = 3t^2+2t`