Revise: Methods of Differentiation

lim(x→a) f(x) = L: f(x) gets closer to L as x gets closer to a.

(x²−1)/(x−1) → limit 2 as x→1.

Try substitution first; if 0/0, factor and cancel, then substitute again.

(x²−4)/(x−2) → limit 4 as x→2.

Average rate = secant slope; instantaneous rate = tangent slope.

h(t)=t², t=0 to 2 → average rate 2.

f'(x) = lim(h→0) [f(x+h)−f(x)]/h.

f(x)=x² → f'(x)=2x.

d/dx[x^n] = n·x^(n−1).

x⁵ → 5x⁴.

Differentiate a polynomial term by term.

3x²+5x−7 → 6x+5.

(f·g)' = f'g + fg'.

x²(x+3) → 3x²+6x.

(f/g)' = (f'g − fg')/g².

x²/(x+1) → (x²+2x)/(x+1)².

Differentiate outer, then multiply by derivative of inner.

(x−2)² → 2(x−2).

Differentiate both sides; every y-term gets an extra dy/dx.

x²+y²=25 at (3,4) → dy/dx=−3/4.

A mixed xy term needs the Product Rule.

x²+xy+y²=7 at (1,2) → dy/dx=−4/5.

A limit can exist even where the function itself is undefined.

(x²−9)/(x−3) → limit 6 as x→3.

0/0 means "try again," not "undefined" — factor first.

(x²−25)/(x−5) → limit 10 as x→5.

Shrinking the secant interval to a point turns it into the tangent.

f(x)=x², x=1 to 4 → average rate 5.

Simplify (factor out h) before substituting h=0.

f(x)=x² → f'(5)=10.

Comes from expanding (x+h)^n with the Binomial Theorem.

4x³ → f'(2)=48.

Follows from the limit of a sum = sum of the limits.

2x³−4x+9 → f'(1)=2.

Never just multiply the two derivatives together.

x(x+2) → f'(3)=8.

Derivable from the Product Rule applied to f·g⁻¹.

x/(x+2) → f'(0)=0.5.

dy/dx = (dy/du)·(du/dx).

(2x−1)³ → f'(1)=6.

Needed whenever y cannot be isolated as y=f(x).

x²+y²=25 at (0,5) → dy/dx=0.

Combines the Power, Product, and Chain Rules in one problem.

x²+xy+y²=7 at (2,1) → dy/dx=−1.25.