Example of using the
limit definition of the derivative to find its formula
f(x) = -2x2-3x-5
1) f(x+h)= -2(x+h)2-3(x+h)-5 (just replace x by x+h where you see an x)
= -2(x2+2hx+h2) -3(x+h)-5 (square the x+h)
= -2x2-4hx-2h2-3x-3h-5 (eliminate the
parentheses)
2) f(x+h)-f(x)= -2x2-4hx-2h2-3x-3h-5-(-2x2-3x-5) (subtract the f(x) expression)
= -2x2-4hx-2h2-3x-3h-5+2x2+3x+5 (eliminate the parentheses)
= -4hx-2h2-3h (simplify…all terms should contain a least one factor of h)
3) [f(x+h)-f(x)]/h =-4x-2h-3 (divide each term by h)
4) lim [f(x+h)-f(x)]/h
h->0 = -4x-3 (sub in 0 for h…all terms with h will disappear)
this is f '(x)=-4x-3 (this is your answer)
Not all derivatives are this easy. Some require
quite a bit of algebra before the division in step 3