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

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 #