Purpose of This Lesson No new material here. This is a full review of every derivative technique from Units 2 and 3. The goal is recognizing which method fits which problem.
Unit 2: Basic Derivative Rules Power rule
d d x [ x n ] = n x n − 1 \frac{d}{dx}[x^n] = nx^{n-1} d x d [ x n ] = n x n − 1 Constant rule (where
c c c is any constant)
d d x [ c ] = 0 \frac{d}{dx}[c] = 0 d x d [ c ] = 0 Constant multiple rule
d d x [ c ⋅ f ( x ) ] = c ⋅ f ′ ( x ) \frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x) d x d [ c ⋅ f ( x )] = c ⋅ f ′ ( x ) Sum and difference rule
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 ) Product rule
d d x [ f ( x ) ⋅ g ( x ) ] = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) \frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x) d x d [ f ( x ) ⋅ g ( x )] = f ′ ( x ) g ( x ) + f ( x ) g ′ ( x ) Quotient rule
d d x [ f ( x ) g ( x ) ] = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) [ g ( x ) ] 2 \frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} d x d [ g ( x ) f ( x ) ] = [ g ( x ) ] 2 f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) Exponential functions
d d x [ e x ] = e x d d x [ a x ] = a x ln ( a ) \frac{d}{dx}[e^x] = e^x \qquad \frac{d}{dx}[a^x] = a^x \ln(a) d x d [ e x ] = e x d x d [ a x ] = a x ln ( a ) Logarithmic functions
d d x [ ln x ] = 1 x d d x [ log a x ] = 1 x ln ( a ) \frac{d}{dx}[\ln x] = \frac{1}{x} \qquad \frac{d}{dx}[\log_a x] = \frac{1}{x \ln(a)} d x d [ ln x ] = x 1 d x d [ log a x ] = x ln ( a ) 1 Trigonometric functions
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 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 Memory aid: functions starting with “c” have a negative sign in front of their derivatives.
Unit 3: Advanced Derivative Techniques Chain rule
For composite functions
f ( g ( x ) ) f(g(x)) f ( g ( x )) :
d d x [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) ⋅ g ′ ( x ) \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) d x d [ f ( g ( x ))] = f ′ ( g ( x )) ⋅ g ′ ( x ) Think: derivative of the outer function (leave the inside alone) times the derivative of the inside function.
The “onion rule”: peel away layers from the outside in.
Implicit differentiation
For equations with mixed
x x x and
y y y terms (like
x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25 ):
Differentiate both sides with respect to x x x Include d y d x \frac{dy}{dx} d x d y y y y Collect all d y d x \frac{dy}{dx} d x d y Factor out d y d x \frac{dy}{dx} d x d y Solve for d y d x \frac{dy}{dx} d x d y Derivative of inverse functions
d d x [ f − 1 ( x ) ] = 1 f ′ ( f − 1 ( x ) ) \frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))} d x d [ f − 1 ( x )] = f ′ ( f − 1 ( x )) 1 At a specific point:
[ f − 1 ] ′ ( a ) = 1 f ′ ( f − 1 ( a ) ) [f^{-1}]'(a) = \frac{1}{f'(f^{-1}(a))} [ f − 1 ] ′ ( a ) = f ′ ( f − 1 ( a )) 1 Inverse trigonometric functions
d d x [ sin − 1 ( x ) ] = 1 1 − x 2 \frac{d}{dx}[\sin^{-1}(x)] = \frac{1}{\sqrt{1-x^2}} d x d [ sin − 1 ( x )] = 1 − x 2 1 d d x [ cos − 1 ( x ) ] = − 1 1 − x 2 \frac{d}{dx}[\cos^{-1}(x)] = -\frac{1}{\sqrt{1-x^2}} d x d [ cos − 1 ( x )] = − 1 − x 2 1 d d x [ tan − 1 ( x ) ] = 1 1 + x 2 \frac{d}{dx}[\tan^{-1}(x)] = \frac{1}{1+x^2} d x d [ tan − 1 ( x )] = 1 + x 2 1 d d x [ cot − 1 ( x ) ] = − 1 1 + x 2 \frac{d}{dx}[\cot^{-1}(x)] = -\frac{1}{1+x^2} d x d [ cot − 1 ( x )] = − 1 + x 2 1 d d x [ sec − 1 ( x ) ] = 1 ∣ x ∣ x 2 − 1 \frac{d}{dx}[\sec^{-1}(x)] = \frac{1}{|x|\sqrt{x^2-1}} d x d [ sec − 1 ( x )] = ∣ x ∣ x 2 − 1 1 d d x [ csc − 1 ( x ) ] = − 1 ∣ x ∣ x 2 − 1 \frac{d}{dx}[\csc^{-1}(x)] = -\frac{1}{|x|\sqrt{x^2-1}} d x d [ csc − 1 ( x )] = − ∣ x ∣ x 2 − 1 1 Memory aids:
- “S” functions have subtraction and square roots
- “Co” functions are the negative versions
- Secant has the absolute value (“oh crap!”)
Problem Recognition Strategy When you’re staring at a derivative problem, ask:
A product of two functions? → product ruleA quotient (fraction)? → quotient ruleA function inside another function? → chain rulex x x and y y y mixed together? → implicit differentiationAn inverse function f − 1 f^{-1} f − 1 ? → inverse function formulaAn inverse trig function? → inverse trig derivativese x e^x e x , ln x \ln x ln x , or trig functions? → the specific function rulesMultiple techniques at once? → combine rules as needed (chain rule plus product rule, for instance)Putting It Together: Worked Examples Start with the outermost structure: the last thing you’d do if you plugged in a number. That tells you which rule to use first. Then handle the pieces inside as they come up. Before you start, also ask whether rewriting would make the problem easier.
Example 1: f ( x ) = x 4 + 3 x 2 f(x) = \dfrac{x^4 + 3}{x^2} f ( x ) = x 2 x 4 + 3 It’s a fraction, but the denominator is a single power of
x x x . Skip the quotient rule and split it up first:
f ( x ) = x 2 + 3 x − 2 ⇒ f ′ ( x ) = 2 x − 6 x − 3 = 2 x − 6 x 3 f(x) = x^2 + 3x^{-2} \quad \Rightarrow \quad f'(x) = 2x - 6x^{-3} = 2x - \frac{6}{x^3} f ( x ) = x 2 + 3 x − 2 ⇒ f ′ ( x ) = 2 x − 6 x − 3 = 2 x − x 3 6 Example 2: g ( x ) = x 2 e 3 x g(x) = x^2 e^{3x} g ( x ) = x 2 e 3 x The outermost operation is multiplication, so start with the product rule. The second factor,
e 3 x e^{3x} e 3 x , needs the chain rule.
g ′ ( x ) = 2 x ⋅ e 3 x + x 2 ⋅ 3 e 3 x = x e 3 x ( 2 + 3 x ) g'(x) = 2x \cdot e^{3x} + x^2 \cdot 3e^{3x} = xe^{3x}(2 + 3x) g ′ ( x ) = 2 x ⋅ e 3 x + x 2 ⋅ 3 e 3 x = x e 3 x ( 2 + 3 x ) Example 3: h ( x ) = [ tan − 1 ( 3 x ) ] 2 h(x) = [\tan^{-1}(3x)]^2 h ( x ) = [ tan − 1 ( 3 x ) ] 2 The outermost operation is the square, so this is a chain rule problem first. Inside that you need the arctangent derivative, and then one more chain rule for the
3 x 3x 3 x .
h ′ ( x ) = 2 tan − 1 ( 3 x ) ⋅ 1 1 + ( 3 x ) 2 ⋅ 3 = 6 tan − 1 ( 3 x ) 1 + 9 x 2 h'(x) = 2\tan^{-1}(3x) \cdot \frac{1}{1 + (3x)^2} \cdot 3 = \frac{6\tan^{-1}(3x)}{1 + 9x^2} h ′ ( x ) = 2 tan − 1 ( 3 x ) ⋅ 1 + ( 3 x ) 2 1 ⋅ 3 = 1 + 9 x 2 6 tan − 1 ( 3 x ) Example 4: k ( x ) = ln ( x 3 x 2 + 1 ) k(x) = \ln\left(\dfrac{x^3}{\sqrt{x^2 + 1}}\right) k ( x ) = ln ( x 2 + 1 x 3 ) for
x > 0 x > 0 x > 0 Going straight in means the chain rule wrapped around the quotient rule. Use log properties to break it apart first:
k ( x ) = 3 ln x − 1 2 ln ( x 2 + 1 ) k(x) = 3\ln x - \frac{1}{2}\ln(x^2 + 1) k ( x ) = 3 ln x − 2 1 ln ( x 2 + 1 ) k ′ ( x ) = 3 x − 1 2 ⋅ 2 x x 2 + 1 = 3 x − x x 2 + 1 k'(x) = \frac{3}{x} - \frac{1}{2} \cdot \frac{2x}{x^2 + 1} = \frac{3}{x} - \frac{x}{x^2 + 1} k ′ ( x ) = x 3 − 2 1 ⋅ x 2 + 1 2 x = x 3 − x 2 + 1 x Example 5: Find
d y d x \frac{dy}{dx} d x d y for
x 2 y + e y = 3 x x^2 y + e^y = 3x x 2 y + e y = 3 x .
x x x and
y y y are mixed together, so use implicit differentiation. The term
x 2 y x^2 y x 2 y is a product, so it needs the product rule, and
e y e^y e y needs the chain rule:
2 x y + x 2 d y d x + e y d y d x = 3 2xy + x^2\frac{dy}{dx} + e^y\frac{dy}{dx} = 3 2 x y + x 2 d x d y + e y d x d y = 3 d y d x ( x 2 + e y ) = 3 − 2 x y ⇒ d y d x = 3 − 2 x y x 2 + e y \frac{dy}{dx}(x^2 + e^y) = 3 - 2xy \quad \Rightarrow \quad \frac{dy}{dx} = \frac{3 - 2xy}{x^2 + e^y} d x d y ( x 2 + e y ) = 3 − 2 x y ⇒ d x d y = x 2 + e y 3 − 2 x y Example 6: Mixed rules from a table
The functions
p p p and
q q q are differentiable, and
q q q is increasing. Use the table to find each value.
(a) A ( x ) = p ( q ( x ) ) A(x) = p(q(x)) A ( x ) = p ( q ( x )) . Find
A ′ ( 1 ) A'(1) A ′ ( 1 ) .
A function inside a function, so use the chain rule. From the table,
q ( 1 ) = 3 q(1) = 3 q ( 1 ) = 3 :
A ′ ( 1 ) = p ′ ( q ( 1 ) ) ⋅ q ′ ( 1 ) = p ′ ( 3 ) ⋅ 2 = 3 ⋅ 2 = 6 A'(1) = p'(q(1)) \cdot q'(1) = p'(3) \cdot 2 = 3 \cdot 2 = 6 A ′ ( 1 ) = p ′ ( q ( 1 )) ⋅ q ′ ( 1 ) = p ′ ( 3 ) ⋅ 2 = 3 ⋅ 2 = 6 (b) B ( x ) = x ⋅ p ( x 2 ) B(x) = x \cdot p(x^2) B ( x ) = x ⋅ p ( x 2 ) . Find
B ′ ( 2 ) B'(2) B ′ ( 2 ) .
The outermost operation is multiplication, so use the product rule, with the chain rule on
p ( x 2 ) p(x^2) p ( x 2 ) :
B ′ ( x ) = p ( x 2 ) + x ⋅ p ′ ( x 2 ) ⋅ 2 x = p ( x 2 ) + 2 x 2 p ′ ( x 2 ) B'(x) = p(x^2) + x \cdot p'(x^2) \cdot 2x = p(x^2) + 2x^2 p'(x^2) B ′ ( x ) = p ( x 2 ) + x ⋅ p ′ ( x 2 ) ⋅ 2 x = p ( x 2 ) + 2 x 2 p ′ ( x 2 ) B ′ ( 2 ) = p ( 4 ) + 8 p ′ ( 4 ) = 3 + 8 ( 2 ) = 19 B'(2) = p(4) + 8p'(4) = 3 + 8(2) = 19 B ′ ( 2 ) = p ( 4 ) + 8 p ′ ( 4 ) = 3 + 8 ( 2 ) = 19 (c) C ( x ) = q − 1 ( x ) C(x) = q^{-1}(x) C ( x ) = q − 1 ( x ) . Find
C ′ ( 6 ) C'(6) C ′ ( 6 ) .
You don’t have a formula for
q − 1 q^{-1} q − 1 , so use the inverse derivative formula. From the table,
q ( 3 ) = 6 q(3) = 6 q ( 3 ) = 6 , so
q − 1 ( 6 ) = 3 q^{-1}(6) = 3 q − 1 ( 6 ) = 3 :
C ′ ( 6 ) = 1 q ′ ( q − 1 ( 6 ) ) = 1 q ′ ( 3 ) = 1 4 C'(6) = \frac{1}{q'(q^{-1}(6))} = \frac{1}{q'(3)} = \frac{1}{4} C ′ ( 6 ) = q ′ ( q − 1 ( 6 )) 1 = q ′ ( 3 ) 1 = 4 1 
