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Differentiation rules

Computing every derivative from first principles is tedious. Instead, we build a small library of rules and a table of standard derivatives. With these, the derivative of almost any algebraic-trigonometric expression follows by mechanical steps.

The five basic rules

Let u(x),v(x)u(x), v(x) be differentiable functions, cc a constant.

(R1) Constant rule. (c)′=0(c)' = 0.

(R2) Constant multiple. (c⋅u)′=c⋅u′(c \cdot u)' = c \cdot u'.

(R3) Sum/difference. (u±v)′=u′±v′(u \pm v)' = u' \pm v'.

(R4) Product rule. (uv)′=u′v+uv′(u v)' = u' v + u v'.

(R5) Quotient rule. (uv)′=u′v−uv′v2\left(\dfrac{u}{v}\right)' = \dfrac{u' v - u v'}{v^2}, where v≠0v \ne 0.

Proofs (sketch)

R4 proof: (u(x+h)v(x+h)−u(x)v(x))/h(u(x+h)v(x+h) - u(x) v(x))/h. Add and subtract u(x)v(x+h)u(x) v(x+h): =[u(x+h)−u(x)]v(x+h)+u(x)[v(x+h)−v(x)]h=u(x+h)−u(x)hv(x+h)+u(x)v(x+h)−v(x)h.= \dfrac{[u(x+h) - u(x)] v(x+h) + u(x)[v(x+h) - v(x)]}{h} = \dfrac{u(x+h) - u(x)}{h} v(x+h) + u(x) \dfrac{v(x+h) - v(x)}{h}. As h→0h \to 0: u′(x)v(x)+u(x)v′(x)u'(x) v(x) + u(x) v'(x). \qed\qed

R5 proof: similar idea, with more bookkeeping. Or derive R5 from R4 plus (1/v)′=−v′/v2(1/v)' = -v'/v^2.

Standard derivatives table

fff′f'
cc (constant)00
xx11
xnx^nnxn−1n x^{n-1}
x\sqrt{x}1/(2x)1/(2\sqrt{x})
1/x1/x−1/x2-1/x^2
sin⁡x\sin xcos⁡x\cos x
cos⁡x\cos x−sin⁡x-\sin x
tan⁡x\tan xsec⁡2x\sec^2 x
cot⁡x\cot x−csc⁡2x-\csc^2 x
sec⁡x\sec xsec⁡xtan⁡x\sec x \tan x
csc⁡x\csc x−csc⁡xcot⁡x-\csc x \cot x
exe^xexe^x
axa^xaxln⁡aa^x \ln a
ln⁡x\ln x1/x1/x
log⁡ax\log_a x1/(xln⁡a)1/(x \ln a)

Worked examples

Example 1. f(x)=5x3−4x+7f(x) = 5 x^3 - 4 x + 7. Find f′(x)f'(x).

By R2, R3: f′(x)=15x2−4f'(x) = 15 x^2 - 4.

Example 2. f(x)=(3x+1)(2x−5)f(x) = (3 x + 1)(2 x - 5). Find f′(x)f'(x).

By R4: f′(x)=3⋅(2x−5)+(3x+1)⋅2=6x−15+6x+2=12x−13f'(x) = 3 \cdot (2 x - 5) + (3 x + 1) \cdot 2 = 6 x - 15 + 6 x + 2 = 12 x - 13.

(Or expand first: f(x)=6x2−13x−5⇒f′(x)=12x−13f(x) = 6 x^2 - 13 x - 5 \Rightarrow f'(x) = 12 x - 13. ✓)

Example 3. f(x)=x2+1x−1f(x) = \dfrac{x^2 + 1}{x - 1}. Find f′(x)f'(x).

By R5: f′=2x⋅(x−1)−(x2+1)⋅1(x−1)2=2x2−2x−x2−1(x−1)2=x2−2x−1(x−1)2f' = \dfrac{2 x \cdot (x - 1) - (x^2 + 1) \cdot 1}{(x - 1)^2} = \dfrac{2 x^2 - 2 x - x^2 - 1}{(x - 1)^2} = \dfrac{x^2 - 2 x - 1}{(x - 1)^2}.

Example 4. f(x)=xsin⁡xf(x) = x \sin x. Find f′(x)f'(x).

By R4: f′(x)=1⋅sin⁡x+x⋅cos⁡x=sin⁡x+xcos⁡xf'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x.

Example 5. f(x)=sin⁡xxf(x) = \dfrac{\sin x}{x}. Find f′(x)f'(x).

By R5: f′(x)=cos⁡x⋅x−sin⁡x⋅1x2=xcos⁡x−sin⁡xx2f'(x) = \dfrac{\cos x \cdot x - \sin x \cdot 1}{x^2} = \dfrac{x \cos x - \sin x}{x^2}.

Try it yourself

Find f′(x)f'(x):

  1. f(x)=7x4−3x2+2f(x) = 7 x^4 - 3 x^2 + 2.
  2. f(x)=x2+3sin⁡xf(x) = x^2 + 3 \sin x.
  3. f(x)=(x2+1)(x−2)f(x) = (x^2 + 1)(x - 2).
  4. f(x)=x3x+1f(x) = \dfrac{x^3}{x + 1}.
  5. f(x)=1+x1−xf(x) = \dfrac{1 + x}{1 - x}.
  6. f(x)=(x+1)(x+2)(x+3)f(x) = (x + 1)(x + 2)(x + 3). (Use R4 twice or expand.)
  7. f(x)=x2cos⁡xf(x) = x^2 \cos x.
  8. f(x)=sin⁡x1+cos⁡xf(x) = \dfrac{\sin x}{1 + \cos x}.
  9. f(x)=x⋅sin⁡xf(x) = \sqrt{x} \cdot \sin x.
  10. f(x)=excos⁡xf(x) = e^x \cos x.
  11. f(x)=ln⁡xxf(x) = \dfrac{\ln x}{x}.
  12. f(x)=11+x2f(x) = \dfrac{1}{1 + x^2}.

Pitfalls / Tricks

  • Product rule: (uv)′≠u′v′(u v)' \ne u' v'. Both terms must be present.
  • Quotient rule: numerator is u′v−uv′u' v - u v' (order matters).
  • For three-factor products, apply the product rule twice: (uvw)′=u′vw+uv′w+uvw′(u v w)' = u' v w + u v' w + u v w'.
  • Insight. Derivatives are linear: (cu+dv)′=cu′+dv′(c u + d v)' = c u' + d v'. Whenever you see a linear combination, just differentiate term by term.

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