Chapter 2: Differentiation — Definition and Basic Derivative Rules# 2.1 Average Rate of Change vs. Instantaneous Rate of Change# Average Rate of Change# The average rate of change of f ( x ) f(x) f ( x ) x = a x=a x = a x = b x=b x = b
f ( b ) − f ( a ) b − a . \frac{f(b) - f(a)}{b - a}. b − a f ( b ) − f ( a ) . This is essentially the slope of the secant line connecting ( a , f ( a ) ) \bigl(a, f(a)\bigr) ( a , f ( a ) ) ( b , f ( b ) ) \bigl(b, f(b)\bigr) ( b , f ( b ) ) It tells us how fast the function’s output changes, on average, as the input goes from a a a b b b Instantaneous Rate of Change# The instantaneous rate of change at x = a x=a x = a f ′ ( a ) f'(a) f ′ ( a ) x = a x=a x = a
f ′ ( a ) = lim h → 0 f ( a + h ) − f ( a ) h . f'(a) \;=\; \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. f ′ ( a ) = h → 0 lim h f ( a + h ) − f ( a ) . This limit expresses how f ( x ) f(x) f ( x ) x = a x=a x = a
2.2 Defining the Derivative of a Function & Derivative Notation# Limit Definition of the Derivative# f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h . f'(x) \;=\; \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. f ′ ( x ) = h → 0 lim h f ( x + h ) − f ( x ) . If this limit exists for each x x x f ′ ( x ) f'(x) f ′ ( x ) function that gives the slope at every point. If the limit does not exist at some x x x f f f Notation# Leibniz Notation : d y d x \frac{dy}{dx} d x d y d d x [ f ( x ) ] \frac{d}{dx}[f(x)] d x d [ f ( x )] Lagrange Notation : f ′ ( x ) f'(x) f ′ ( x ) Newton’s Notation : y ˙ \dot{y} y ˙ You might see all of these. They convey the same concept: the instantaneous rate of change of y y y x x x
2.3 Estimating Derivatives of a Function at a Point# Sometimes you only have a graph or a table rather than an explicit formula. To approximate f ′ ( a ) f'(a) f ′ ( a )
From a Table : Use the difference quotient f ( x 2 ) − f ( x 1 ) x 2 − x 1 \frac{f(x_2)-f(x_1)}{x_2-x_1} x 2 − x 1 f ( x 2 ) − f ( x 1 ) x 1 x_1 x 1 x 2 x_2 x 2 a a a x 1 , x 2 x_1, x_2 x 1 , x 2 a a a From a Graph : Draw a tangent line at x = a x=a x = a 2.4 Connecting Differentiability and Continuity# If f f f differentiable at x = a x=a x = a f f f continuous at x = a x=a x = a However, a function can be continuous but not differentiable if it has a “sharp corner,” “vertical tangent,” or a cusp. For instance, ∣ x ∣ |x| ∣ x ∣ x = 0 x=0 x = 0 2.5 The Power Rule# A direct result of applying the limit definition to power functions is the Power Rule :
d d x [ x n ] = n x n − 1 . \frac{d}{dx}[x^n] = n\,x^{n-1}. d x d [ x n ] = n x n − 1 . Examples:
d d x [ x 3 ] = 3 x 2 \frac{d}{dx}[x^3] = 3x^2 d x d [ x 3 ] = 3 x 2 d d x [ x ] = 1 \frac{d}{dx}[x] = 1 d x d [ x ] = 1 d d x [ x ] = d d x [ x 1 / 2 ] = 1 2 x − 1 / 2 = 1 2 x . \frac{d}{dx}[\sqrt{x}] = \frac{d}{dx}[x^{1/2}] = \tfrac12 x^{-1/2} = \frac{1}{2\sqrt{x}}. d x d [ x ] = d x d [ x 1/2 ] = 2 1 x − 1/2 = 2 x 1 . 2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple# Constant Function Rule d d x [ c ] = 0 , \frac{d}{dx}[c] = 0, d x d [ c ] = 0 , c c c
Sum / Difference Rules
d d x [ f ( x ) ± g ( x ) ] = f ′ ( x ) ± g ′ ( x ) . \frac{d}{dx}[f(x)\pm g(x)] \;=\; f'(x)\;\pm\;g'(x). d x d [ f ( x ) ± g ( x )] = f ′ ( x ) ± g ′ ( x ) . Constant Multiple Rule
d d x [ c f ( x ) ] = c f ′ ( x ) . \frac{d}{dx}[c\,f(x)] \;=\; c\,f'(x). d x d [ c f ( x )] = c f ′ ( x ) . All these rules ensure derivatives behave in a linear way with respect to addition and scalar multiplication.
2.7 Derivatives of sin x \sin x sin x cos x \cos x cos x e x e^x e x ln x \ln x ln x # sin x \sin x sin x cos x \cos x cos x
d d x [ sin x ] = cos x \frac{d}{dx}[\sin x] = \cos x d x d [ sin x ] = cos x d d x [ cos x ] = − sin x \frac{d}{dx}[\cos x] = -\,\sin x d x d [ cos x ] = − sin x e x e^x e x d d x [ e x ] = e x . \frac{d}{dx}[e^x] = e^x. d x d [ e x ] = e x .
ln x \ln x ln x d d x [ ln x ] = 1 x , x > 0. \frac{d}{dx}[\ln x] = \frac{1}{x},\quad x>0. d x d [ ln x ] = x 1 , x > 0.
These are standard derivatives you should memorize. They can be derived rigorously using the limit definition, but in practice they’re typically treated as fundamental facts.
2.8 The Product Rule# When you have f ( x ) = u ( x ) v ( x ) f(x) = u(x)\,v(x) f ( x ) = u ( x ) v ( x )
f ′ ( x ) = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) . f'(x) = u'(x)\,v(x) \;+\; u(x)\,v'(x). f ′ ( x ) = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) . Example : If f ( x ) = ( x 3 ) ( sin x ) f(x)= (x^3)\,(\sin x) f ( x ) = ( x 3 ) ( sin x )
f ′ ( x ) = ( 3 x 2 ) ( sin x ) + ( x 3 ) ( cos x ) . f'(x)= (3x^2)(\sin x) \;+\; (x^3)(\cos x). f ′ ( x ) = ( 3 x 2 ) ( sin x ) + ( x 3 ) ( cos x ) . 2.9 The Quotient Rule# For a function that’s the ratio of two differentiable functions u ( x ) v ( x ) \frac{u(x)}{v(x)} v ( x ) u ( x )
( u v ) ′ = u ′ ( x ) v ( x ) − u ( x ) v ′ ( x ) ( v ( x ) ) 2 . \left(\frac{u}{v}\right)' = \frac{u'(x)\,v(x) - u(x)\,v'(x)}{(\,v(x)\,)^2}. ( v u ) ′ = ( v ( x ) ) 2 u ′ ( x ) v ( x ) − u ( x ) v ′ ( x ) . Example : If g ( x ) = 2 x 1 + x 2 g(x)= \frac{2x}{1+x^2} g ( x ) = 1 + x 2 2 x
g ′ ( x ) = ( 2 ) ( 1 + x 2 ) − ( 2 x ) ( 2 x ) ( 1 + x 2 ) 2 = 2 ( 1 + x 2 ) − 4 x 2 ( 1 + x 2 ) 2 . g'(x)= \frac{(2)(1+x^2) - (2x)(2x)}{(1+x^2)^2} = \frac{2(1+x^2) -4x^2}{(1+x^2)^2}. g ′ ( x ) = ( 1 + x 2 ) 2 ( 2 ) ( 1 + x 2 ) − ( 2 x ) ( 2 x ) = ( 1 + x 2 ) 2 2 ( 1 + x 2 ) − 4 x 2 . 2.10 The Derivatives of tan x \tan x tan x cot x \cot x cot x sec x \sec x sec x csc x \csc x csc x # Using the quotient rule (or known identities), we get:
d d x [ tan x ] = sec 2 x . \frac{d}{dx}[\tan x] = \sec^2 x. d x d [ tan x ] = sec 2 x . d d x [ cot x ] = − csc 2 x . \frac{d}{dx}[\cot x] = -\,\csc^2 x. d x d [ cot x ] = − csc 2 x . d d x [ sec x ] = sec x tan x . \frac{d}{dx}[\sec x] = \sec x\,\tan x. d x d [ sec x ] = sec x tan x . d d x [ csc x ] = − csc x cot x . \frac{d}{dx}[\csc x] = -\,\csc x\,\cot x. d x d [ csc x ] = − csc x cot x . These can also be memorized or derived systematically using tan x = sin x cos x \tan x= \frac{\sin x}{\cos x} tan x = c o s x s i n x
Self-Check Practice# Below are several sample problems to reinforce your understanding. Try each before looking at the solution.
Self-Check 1# Problem : Using the limit definition of the derivative, show that:
d d x [ x 2 ] = 2 x . \frac{d}{dx}[x^2]\;=\;2x. d x d [ x 2 ] = 2 x . Solution 1 # By definition:
d d x [ x 2 ] = lim h → 0 ( x + h ) 2 − x 2 h . \frac{d}{dx}[x^2] \;=\; \lim_{h\to 0}\frac{(x+h)^2 -x^2}{h}. d x d [ x 2 ] = h → 0 lim h ( x + h ) 2 − x 2 . Expand the numerator: ( x + h ) 2 = x 2 + 2 x h + h 2 . (x+h)^2 = x^2 +2xh +h^2. ( x + h ) 2 = x 2 + 2 x h + h 2 .
So,
( x + h ) 2 − x 2 h = x 2 + 2 x h + h 2 − x 2 h = 2 x h + h 2 h . \frac{(x+h)^2 -x^2}{h} = \frac{x^2 +2xh +h^2 -x^2}{h} = \frac{2xh + h^2}{h}. h ( x + h ) 2 − x 2 = h x 2 + 2 x h + h 2 − x 2 = h 2 x h + h 2 . Factor out h h h
2 x h + h 2 h = 2 x + h . \frac{2xh +h^2}{h} = 2x +h. h 2 x h + h 2 = 2 x + h . Now take the limit as h → 0 h\to 0 h → 0 2 x + h → 2 x 2x +h \to 2x 2 x + h → 2 x
Hence, d d x [ x 2 ] = 2 x \frac{d}{dx}[x^2] = \boxed{2x} d x d [ x 2 ] = 2 x
Self-Check 2# Problem : Differentiate f ( x ) = 7 x 4 − 3 x + 10 f(x)= 7x^4 -3x +10 f ( x ) = 7 x 4 − 3 x + 10 Power, Sum, and Constant Rules .
Solution 2 # Term by term:d d x [ 7 x 4 ] = 7 ⋅ 4 x 3 = 28 x 3 . \frac{d}{dx}[7x^4] = 7\cdot 4x^{3}=28x^3. d x d [ 7 x 4 ] = 7 ⋅ 4 x 3 = 28 x 3 . d d x [ − 3 x ] = − 3. \frac{d}{dx}[-3x] = -3. d x d [ − 3 x ] = − 3. d d x [ 10 ] = 0. \frac{d}{dx}[10] = 0. d x d [ 10 ] = 0. Putting it all together:
f ′ ( x ) = 28 x 3 − 3. f'(x)= 28x^3 -3. f ′ ( x ) = 28 x 3 − 3. Self-Check 3# Problem : Differentiate g ( x ) = x 5 ln x g(x)= x^5 \ln x g ( x ) = x 5 ln x Product Rule . Assume x > 0 x>0 x > 0
Solution 3 # Let u ( x ) = x 5 u(x)= x^5 u ( x ) = x 5 v ( x ) = ln x v(x)= \ln x v ( x ) = ln x
Then u ′ ( x ) = 5 x 4 u'(x)= 5x^4 u ′ ( x ) = 5 x 4 v ′ ( x ) = 1 x v'(x)= \frac{1}{x} v ′ ( x ) = x 1
Product Rule:
g ′ ( x ) = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) = ( 5 x 4 ) ( ln x ) + ( x 5 ) ( 1 x ) . g'(x)= u'(x)\,v(x) + u(x)\,v'(x) = (5x^4)(\ln x) + (x^5)\left(\frac{1}{x}\right). g ′ ( x ) = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) = ( 5 x 4 ) ( ln x ) + ( x 5 ) ( x 1 ) . Simplify the second term:
( x 5 ) ( 1 x ) = x 4 . (x^5)\left(\frac{1}{x}\right) = x^4. ( x 5 ) ( x 1 ) = x 4 . Therefore,
g ′ ( x ) = 5 x 4 ln x + x 4 = x 4 ( 5 ln x + 1 ) . g'(x)= 5x^4 \ln x + x^4 = x^4(5\ln x + 1). g ′ ( x ) = 5 x 4 ln x + x 4 = x 4 ( 5 ln x + 1 ) . Self-Check 4# Problem : Differentiate h ( x ) = sin x x 2 h(x)= \frac{\sin x}{x^2} h ( x ) = x 2 s i n x Quotient Rule .
Solution 4 # Let u ( x ) = sin x u(x)= \sin x u ( x ) = sin x v ( x ) = x 2 v(x)= x^2 v ( x ) = x 2
Then u ′ ( x ) = cos x u'(x)= \cos x u ′ ( x ) = cos x v ′ ( x ) = 2 x v'(x)= 2x v ′ ( x ) = 2 x
Quotient Rule:
h ′ ( x ) = u ′ ( x ) v ( x ) − u ( x ) v ′ ( x ) ( v ( x ) ) 2 = ( cos x ) ( x 2 ) − ( sin x ) ( 2 x ) ( x 2 ) 2 . h'(x)= \frac{u'(x)\,v(x) - u(x)\,v'(x)}{(v(x))^2} = \frac{(\cos x)(x^2) -(\sin x)(2x)}{(x^2)^2}. h ′ ( x ) = ( v ( x ) ) 2 u ′ ( x ) v ( x ) − u ( x ) v ′ ( x ) = ( x 2 ) 2 ( cos x ) ( x 2 ) − ( sin x ) ( 2 x ) . Simplify numerator and denominator:
= x 2 cos x − 2 x sin x x 4 = x 2 cos x x 4 − 2 x sin x x 4 = cos x x 2 − 2 sin x x 3 . = \frac{x^2 \cos x -2x \sin x}{x^4} = \frac{x^2\cos x}{x^4} - \frac{2x\sin x}{x^4} = \frac{\cos x}{x^2} - \frac{2\sin x}{x^3}. = x 4 x 2 cos x − 2 x sin x = x 4 x 2 cos x − x 4 2 x sin x = x 2 cos x − x 3 2 sin x . Final derivative:
h ′ ( x ) = cos x x 2 − 2 sin x x 3 . h'(x)= \boxed{\frac{\cos x}{x^2} - \frac{2\sin x}{x^3}}. h ′ ( x ) = x 2 cos x − x 3 2 sin x . Self-Check 5# Problem : Find the derivatives of each of the following trigonometric functions:
p ( x ) = tan x p(x)= \tan x p ( x ) = tan x q ( x ) = sec x q(x)= \sec x q ( x ) = sec x Solution 5 # p ( x ) = tan x p(x)= \tan x p ( x ) = tan x
So, d d x [ tan x ] = sec 2 x . \frac{d}{dx}[\tan x]= \sec^2 x. d x d [ tan x ] = sec 2 x .
q ( x ) = sec x q(x)= \sec x q ( x ) = sec x
Alternatively, you can do d d x [ sec x ] = sec x tan x \frac{d}{dx}[\sec x]= \sec x \tan x d x d [ sec x ] = sec x tan x
Hence, the results:
d d x [ tan x ] = sec 2 x and d d x [ sec x ] = sec x tan x . \frac{d}{dx}[\tan x]= \sec^2 x \quad\text{and}\quad \frac{d}{dx}[\sec x]= \sec x \tan x. d x d [ tan x ] = sec 2 x and d x d [ sec x ] = sec x tan x . 
