This section proves six standard derivative formulas directly from the first-principles limit of the previous section. Each proof follows the same three-step pattern: write f ( x + h ) f(x+h) f ( x + h ) f ( x + h ) − f ( x ) h \dfrac{f(x+h)-f(x)}{h} h f ( x + h ) − f ( x ) h → 0 h\to0 h → 0
(1) Derivative of x n x^n x n n ∈ N n\in N n ∈ N Let f ( x ) = x n f(x)=x^n f ( x ) = x n f ( x + h ) = ( x + h ) n f(x+h)=(x+h)^n f ( x + h ) = ( x + h ) n
( x + h ) n = x n + n C 1 x n − 1 h + n C 2 x n − 2 h 2 + ⋯ + h n (x+h)^n=x^n+{}^nC_1x^{n-1}h+{}^nC_2x^{n-2}h^2+\cdots+h^n ( x + h ) n = x n + n C 1 x n − 1 h + n C 2 x n − 2 h 2 + ⋯ + h n
so
f ( x + h ) − f ( x ) h = n C 1 x n − 1 h + n C 2 x n − 2 h 2 + ⋯ + h n h = n C 1 x n − 1 + n C 2 x n − 2 h + ⋯ + h n − 1 \dfrac{f(x+h)-f(x)}{h}=\dfrac{{}^nC_1x^{n-1}h+{}^nC_2x^{n-2}h^2+\cdots+h^n}{h}={}^nC_1x^{n-1}+{}^nC_2x^{n-2}h+\cdots+h^{n-1} h f ( x + h ) − f ( x ) = h n C 1 x n − 1 h + n C 2 x n − 2 h 2 + ⋯ + h n = n C 1 x n − 1 + n C 2 x n − 2 h + ⋯ + h n − 1
Every term after the first carries at least one positive power of h h h h → 0 h\to0 h → 0 h ≠ 0 h\ne0 h = 0 n C 1 x n − 1 = n x n − 1 {}^nC_1x^{n-1}=nx^{n-1} n C 1 x n − 1 = n x n − 1
f ( x ) = x n ⇒ f ′ ( x ) = n x n − 1 \boxed{f(x)=x^n\ \Rightarrow\ f'(x)=nx^{n-1}} f ( x ) = x n ⇒ f ′ ( x ) = n x n − 1
(2) Derivative of sin x \sin x sin x Let f ( x ) = sin x f(x)=\sin x f ( x ) = sin x sin A − sin B = 2 cos ( A + B 2 ) sin ( A − B 2 ) \sin A-\sin B=2\cos\!\left(\dfrac{A+B}2\right)\sin\!\left(\dfrac{A-B}2\right) sin A − sin B = 2 cos ( 2 A + B ) sin ( 2 A − B ) A = x + h , B = x A=x+h,B=x A = x + h , B = x
f ( x + h ) − f ( x ) = sin ( x + h ) − sin x = 2 cos ( x + h 2 ) sin h 2 f(x+h)-f(x)=\sin(x+h)-\sin x=2\cos\!\left(x+\dfrac h2\right)\sin\dfrac h2 f ( x + h ) − f ( x ) = sin ( x + h ) − sin x = 2 cos ( x + 2 h ) sin 2 h
so
f ( x + h ) − f ( x ) h = cos ( x + h 2 ) ⋅ sin ( h / 2 ) h / 2 \dfrac{f(x+h)-f(x)}{h}=\cos\!\left(x+\dfrac h2\right)\cdot\dfrac{\sin(h/2)}{h/2} h f ( x + h ) − f ( x ) = cos ( x + 2 h ) ⋅ h /2 sin ( h /2 )
As h → 0 h\to0 h → 0 cos ( x + h 2 ) → cos x \cos\!\left(x+\dfrac h2\right)\to\cos x cos ( x + 2 h ) → cos x lim θ → 0 sin θ θ = 1 \lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1 lim θ → 0 θ sin θ = 1 → 1 \to1 → 1
f ( x ) = sin x ⇒ f ′ ( x ) = cos x \boxed{f(x)=\sin x\ \Rightarrow\ f'(x)=\cos x} f ( x ) = sin x ⇒ f ′ ( x ) = cos x
(3) Derivative of tan x \tan x tan x Let f ( x ) = tan x = sin x cos x f(x)=\tan x=\dfrac{\sin x}{\cos x} f ( x ) = tan x = cos x sin x
f ( x + h ) − f ( x ) = sin ( x + h ) cos ( x + h ) − sin x cos x = sin ( x + h ) cos x − cos ( x + h ) sin x cos ( x + h ) cos x = sin h cos ( x + h ) cos x f(x+h)-f(x)=\dfrac{\sin(x+h)}{\cos(x+h)}-\dfrac{\sin x}{\cos x}=\dfrac{\sin(x+h)\cos x-\cos(x+h)\sin x}{\cos(x+h)\cos x}=\dfrac{\sin h}{\cos(x+h)\cos x} f ( x + h ) − f ( x ) = cos ( x + h ) sin ( x + h ) − cos x sin x = cos ( x + h ) cos x sin ( x + h ) cos x − cos ( x + h ) sin x = cos ( x + h ) cos x sin h
(using sin ( A − B ) = sin A cos B − cos A sin B \sin(A-B)=\sin A\cos B-\cos A\sin B sin ( A − B ) = sin A cos B − cos A sin B A = x + h , B = x A=x+h,B=x A = x + h , B = x sin h \sin h sin h h h h
f ( x + h ) − f ( x ) h = sin h h ⋅ 1 cos ( x + h ) cos x \dfrac{f(x+h)-f(x)}{h}=\dfrac{\sin h}{h}\cdot\dfrac{1}{\cos(x+h)\cos x} h f ( x + h ) − f ( x ) = h sin h ⋅ cos ( x + h ) cos x 1
As h → 0 h\to0 h → 0 sin h h → 1 \dfrac{\sin h}{h}\to1 h sin h → 1 cos ( x + h ) → cos x \cos(x+h)\to\cos x cos ( x + h ) → cos x 1 cos 2 x \dfrac{1}{\cos^2x} cos 2 x 1
f ( x ) = tan x ⇒ f ′ ( x ) = sec 2 x \boxed{f(x)=\tan x\ \Rightarrow\ f'(x)=\sec^2x} f ( x ) = tan x ⇒ f ′ ( x ) = sec 2 x
(4) Derivative of sec x \sec x sec x Let f ( x ) = sec x = 1 cos x f(x)=\sec x=\dfrac1{\cos x} f ( x ) = sec x = cos x 1
f ( x + h ) − f ( x ) = 1 cos ( x + h ) − 1 cos x = cos x − cos ( x + h ) cos ( x + h ) cos x f(x+h)-f(x)=\dfrac1{\cos(x+h)}-\dfrac1{\cos x}=\dfrac{\cos x-\cos(x+h)}{\cos(x+h)\cos x} f ( x + h ) − f ( x ) = cos ( x + h ) 1 − cos x 1 = cos ( x + h ) cos x cos x − cos ( x + h )
Using cos B − cos A = 2 sin ( A + B 2 ) sin ( A − B 2 ) \cos B-\cos A=2\sin\!\left(\dfrac{A+B}2\right)\sin\!\left(\dfrac{A-B}2\right) cos B − cos A = 2 sin ( 2 A + B ) sin ( 2 A − B ) A = x + h , B = x A=x+h,B=x A = x + h , B = x 2 sin ( x + h 2 ) sin h 2 2\sin\!\left(x+\dfrac h2\right)\sin\dfrac h2 2 sin ( x + 2 h ) sin 2 h
f ( x + h ) − f ( x ) h = sin ( h / 2 ) h / 2 ⋅ sin ( x + h 2 ) cos ( x + h ) cos x \dfrac{f(x+h)-f(x)}{h}=\dfrac{\sin(h/2)}{h/2}\cdot\dfrac{\sin\!\left(x+\frac h2\right)}{\cos(x+h)\cos x} h f ( x + h ) − f ( x ) = h /2 sin ( h /2 ) ⋅ cos ( x + h ) cos x sin ( x + 2 h )
As h → 0 h\to0 h → 0 → 1 \to1 → 1 → sin x cos 2 x = sin x cos x ⋅ 1 cos x \to\dfrac{\sin x}{\cos^2x}=\dfrac{\sin x}{\cos x}\cdot\dfrac1{\cos x} → cos 2 x sin x = cos x sin x ⋅ cos x 1
f ( x ) = sec x ⇒ f ′ ( x ) = sec x tan x \boxed{f(x)=\sec x\ \Rightarrow\ f'(x)=\sec x\tan x} f ( x ) = sec x ⇒ f ′ ( x ) = sec x tan x
(5) Derivative of log x \log x log x x > 0 x>0 x > 0 Let f ( x ) = log x f(x)=\log x f ( x ) = log x
f ( x + h ) − f ( x ) = log ( x + h ) − log x = log ( x + h x ) = log ( 1 + h x ) f(x+h)-f(x)=\log(x+h)-\log x=\log\!\left(\dfrac{x+h}{x}\right)=\log\!\left(1+\dfrac hx\right) f ( x + h ) − f ( x ) = log ( x + h ) − log x = log ( x x + h ) = log ( 1 + x h )
so
f ( x + h ) − f ( x ) h = 1 h log ( 1 + h x ) = 1 x ⋅ log ( 1 + h / x ) h / x \dfrac{f(x+h)-f(x)}{h}=\dfrac1h\log\!\left(1+\dfrac hx\right)=\dfrac1x\cdot\dfrac{\log(1+h/x)}{h/x} h f ( x + h ) − f ( x ) = h 1 log ( 1 + x h ) = x 1 ⋅ h / x log ( 1 + h / x )
As h → 0 h\to0 h → 0 h / x → 0 h/x\to0 h / x → 0 lim t → 0 log ( 1 + t ) t = 1 \lim_{t\to0}\dfrac{\log(1+t)}{t}=1 lim t → 0 t log ( 1 + t ) = 1 → 1 \to1 → 1
f ( x ) = log x ⇒ f ′ ( x ) = 1 x \boxed{f(x)=\log x\ \Rightarrow\ f'(x)=\dfrac1x} f ( x ) = log x ⇒ f ′ ( x ) = x 1
(6) Derivative of a x a^x a x a > 0 a>0 a > 0 Let f ( x ) = a x f(x)=a^x f ( x ) = a x
f ( x + h ) − f ( x ) = a x + h − a x = a x ( a h − 1 ) f(x+h)-f(x)=a^{x+h}-a^x=a^x(a^h-1) f ( x + h ) − f ( x ) = a x + h − a x = a x ( a h − 1 )
so
f ( x + h ) − f ( x ) h = a x ⋅ a h − 1 h \dfrac{f(x+h)-f(x)}{h}=a^x\cdot\dfrac{a^h-1}{h} h f ( x + h ) − f ( x ) = a x ⋅ h a h − 1
As h → 0 h\to0 h → 0 lim h → 0 a h − 1 h = log a \lim_{h\to0}\dfrac{a^h-1}{h}=\log a lim h → 0 h a h − 1 = log a
f ( x ) = a x ⇒ f ′ ( x ) = a x log a \boxed{f(x)=a^x\ \Rightarrow\ f'(x)=a^x\log a} f ( x ) = a x ⇒ f ′ ( x ) = a x log a
Try the following (companion practice box). Using the very same first-principles technique, the text asks the reader to establish five further standard results: (1) f ( x ) = 1 x n ⇒ f ′ ( x ) = − n x n + 1 f(x)=\dfrac1{x^n}\Rightarrow f'(x)=-\dfrac n{x^{n+1}} f ( x ) = x n 1 ⇒ f ′ ( x ) = − x n + 1 n f ( x ) = cos x ⇒ f ′ ( x ) = − sin x f(x)=\cos x\Rightarrow f'(x)=-\sin x f ( x ) = cos x ⇒ f ′ ( x ) = − sin x f ( x ) = cot x ⇒ f ′ ( x ) = − cosec 2 x f(x)=\cot x\Rightarrow f'(x)=-\text{cosec}^2x f ( x ) = cot x ⇒ f ′ ( x ) = − cosec 2 x f ( x ) = cosec x ⇒ f ′ ( x ) = − cosec x cot x f(x)=\text{cosec}\,x\Rightarrow f'(x)=-\text{cosec}\,x\cot x f ( x ) = cosec x ⇒ f ′ ( x ) = − cosec x cot x f ( x ) = e x ⇒ f ′ ( x ) = e x f(x)=e^x\Rightarrow f'(x)=e^x f ( x ) = e x ⇒ f ′ ( x ) = e x f ( x + h ) − f ( x ) f(x+h)-f(x) f ( x + h ) − f ( x ) h h h lim θ → 0 sin θ θ = 1 \lim_{\theta\to0}\frac{\sin\theta}{\theta}=1 lim θ → 0 θ s i n θ = 1 lim h → 0 e h − 1 h = 1 \lim_{h\to0}\frac{e^h-1}{h}=1 lim h → 0 h e h − 1 = 1
Worked examples (first-principles derivatives).
Example: x \sqrt{x} x Let f ( x ) = x f(x)=\sqrt x f ( x ) = x f ( x + h ) = x + h f(x+h)=\sqrt{x+h} f ( x + h ) = x + h
f ( x + h ) − f ( x ) h = x + h − x h \dfrac{f(x+h)-f(x)}{h}=\dfrac{\sqrt{x+h}-\sqrt x}{h} h f ( x + h ) − f ( x ) = h x + h − x
Rationalise by multiplying and dividing by x + h + x \sqrt{x+h}+\sqrt x x + h + x
= ( x + h ) − x h ( x + h + x ) = 1 x + h + x =\dfrac{(x+h)-x}{h\left(\sqrt{x+h}+\sqrt x\right)}=\dfrac{1}{\sqrt{x+h}+\sqrt x} = h ( x + h + x ) ( x + h ) − x = x + h + x 1
As h → 0 h\to0 h → 0 → 1 2 x \to\dfrac1{2\sqrt x} → 2 x 1 f ′ ( x ) = 1 2 x f'(x)=\dfrac1{2\sqrt x} f ′ ( x ) = 2 x 1
Example: cos ( 2 x + 3 ) \cos(2x+3) cos ( 2 x + 3 ) Let f ( x ) = cos ( 2 x + 3 ) f(x)=\cos(2x+3) f ( x ) = cos ( 2 x + 3 ) f ( x + h ) = cos ( 2 x + 2 h + 3 ) = cos [ ( 2 x + 3 ) + 2 h ] f(x+h)=\cos(2x+2h+3)=\cos\big[(2x+3)+2h\big] f ( x + h ) = cos ( 2 x + 2 h + 3 ) = cos [ ( 2 x + 3 ) + 2 h ]
f ( x + h ) − f ( x ) = cos [ ( 2 x + 3 ) + 2 h ] − cos ( 2 x + 3 ) = − 2 sin ( 2 x + 3 + h ) sin h f(x+h)-f(x)=\cos\big[(2x+3)+2h\big]-\cos(2x+3)=-2\sin(2x+3+h)\sin h f ( x + h ) − f ( x ) = cos [ ( 2 x + 3 ) + 2 h ] − cos ( 2 x + 3 ) = − 2 sin ( 2 x + 3 + h ) sin h
f ( x + h ) − f ( x ) h = − 2 sin ( 2 x + 3 + h ) ⋅ sin h h \dfrac{f(x+h)-f(x)}{h}=-2\sin(2x+3+h)\cdot\dfrac{\sin h}{h} h f ( x + h ) − f ( x ) = − 2 sin ( 2 x + 3 + h ) ⋅ h sin h
As h → 0 h\to0 h → 0 → − 2 sin ( 2 x + 3 ) \to-2\sin(2x+3) → − 2 sin ( 2 x + 3 ) f ′ ( x ) = − 2 sin ( 2 x + 3 ) f'(x)=-2\sin(2x+3) f ′ ( x ) = − 2 sin ( 2 x + 3 )