Understand the idea Use the power rule on powers, including negative and fractional powers where defined. Differentiate constants to zero and distribute differentiation through finite sums. Memorize derivatives of the six trigonometric functions and the natural exponential and logarithm. Trigonometric formulas assume radians; domain restrictions remain in force after differentiating.
d d x x n = n x n − 1 , ( e x ) ′ = e x , ( ln x ) ′ = 1 x , ( sin x ) ′ = cos x , ( cos x ) ′ = − sin x , ( tan x ) ′ = sec 2 x \frac d{dx}x^n=nx^{n-1},\quad (e^x)'=e^x,\quad(\ln x)'=\frac1x,\quad(\sin x)'=\cos x,\quad(\cos x)'=-\sin x,\quad(\tan x)'=\sec^2x d x d x n = n x n − 1 , ( e x ) ′ = e x , ( ln x ) ′ = x 1 , ( sin x ) ′ = cos x , ( cos x ) ′ = − sin x , ( tan x ) ′ = sec 2 x Worked example Find the derivative of f ( x ) = 3 x 4 − 2 / x + 5 sin x \displaystyle f(x)=3x^4-2/x+5\sin x f ( x ) = 3 x 4 − 2/ x + 5 sin x
Rewrite
− 2 / x = − 2 x − 1 \displaystyle -2/x=-2x^{-1} − 2/ x = − 2 x − 1 . Differentiate term by term to obtain
f ′ ( x ) = 12 x 3 + 2 x − 2 + 5 cos x \displaystyle f'(x)=12x^3+2x^{-2}+5\cos x f ′ ( x ) = 12 x 3 + 2 x − 2 + 5 cos x , on the original domain
x ≠ 0 \displaystyle x\ne0 x = 0 .
Check this before moving on The remaining trig rules are ( cot x ) ′ = − csc 2 x \displaystyle (\cot x)'=-\csc^2x ( cot x ) ′ = − csc 2 x ( sec x ) ′ = sec x tan x \displaystyle (\sec x)'=\sec x\tan x ( sec x ) ′ = sec x tan x ( csc x ) ′ = − csc x cot x \displaystyle (\csc x)'=-\csc x\cot x ( csc x ) ′ = − csc x cot x
Try it yourself Differentiate g ( x ) = sec x + cot x + csc x \displaystyle g(x)=\sec x+\cot x+\csc x g ( x ) = sec x + cot x + csc x
Show solution and reasoning g ′ ( x ) = sec x tan x − csc 2 x − csc x cot x \displaystyle g'(x)=\sec x\tan x-\csc^2x-\csc x\cot x g ′ ( x ) = sec x tan x − csc 2 x − csc x cot x
Work through AP Calculus AB with a tutor Detailed notes are discussed during the course with worked examples, guided practice, and feedback on mathematical reasoning.
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