The derivative from first principles uses the limit definition f′(x)=limh→0hf(x+h)−f(x). For (i) the derivative is 3x2, for (ii) it is 2x−3, for (iii) it is −x32, and for (iv) it is −(x−1)22.
The first principle of differentiation — also called the ab initio method — is the very definition of a derivative. It measures the instantaneous rate of change by taking the slope of the secant line between two points and letting the distance between them shrink to zero. The formula is:
f′(x)=limh→0hf(x+h)−f(x)
This is not a shortcut rule; it is the reason behind all derivative rules. Every time you use the power rule, product rule, or quotient rule, you are standing on the shoulders of this limit. Let's apply it to each function.
(i) f(x)=x3−27
Step 1: Write f(x+h).
Replace x with x+h:
f(x+h)=(x+h)3−27
Expand (x+h)3 using the binomial theorem:
(x+h)3=x3+3x2h+3xh2+h3
So:
f(x+h)=x3+3x2h+3xh2+h3−27
Step 2: Form the difference quotient.
hf(x+h)−f(x)=h(x3+3x2h+3xh2+h3−27)−(x3−27)
Notice x3 and −27 cancel:
=h3x2h+3xh2+h3
Step 3: Simplify by factoring h.
=hh(3x2+3xh+h2)=3x2+3xh+h2
Step 4: Take the limit as h→0.
f′(x)=limh→0(3x2+3xh+h2)=3x2+0+0=3x2
A common mistake is forgetting to cancel the h before taking the limit. If you try to plug h=0 directly into 00, you get an indeterminate form. Always simplify first.
Notice that the constant −27 disappears in the derivative. Constants always have zero derivative — this is why.
(ii) f(x)=(x−1)(x−2)
Step 1: Expand first for convenience.
f(x)=(x−1)(x−2)=x2−3x+2
Now find f(x+h):
f(x+h)=(x+h)2−3(x+h)+2=x2+2xh+h2−3x−3h+2
Step 2: Difference quotient.
hf(x+h)−f(x)=h(x2+2xh+h2−3x−3h+2)−(x2−3x+2)
Cancel x2, −3x, and 2:
=h2xh+h2−3h
Step 3: Factor h.
=hh(2x+h−3)=2x+h−3
Step 4: Limit.
f′(x)=limh→0(2x+h−3)=2x−3
You could also use the product rule later, but from first principles, expanding first avoids dealing with two separate terms in the numerator.
(iii) f(x)=x21
Step 1: Write f(x+h).
f(x+h)=(x+h)21
Step 2: Difference quotient.
hf(x+h)−f(x)=h(x+h)21−x21
Combine the numerator over a common denominator:
=hx2(x+h)2x2−(x+h)2=h⋅x2(x+h)2x2−(x2+2xh+h2)
Simplify the numerator:
=h⋅x2(x+h)2−2xh−h2
Step 3: Factor h.
=h⋅x2(x+h)2h(−2x−h)=x2(x+h)2−2x−h
Step 4: Limit.
f′(x)=limh→0x2(x+h)2−2x−h=x2(x+0)2−2x−0=x2⋅x2−2x=−x32 …