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) be differentiable functions, c a constant.
(R1) Constant rule.(c)′=0.
(R2) Constant multiple.(c⋅u)′=c⋅u′.
(R3) Sum/difference.(u±v)′=u′±v′.
(R4) Product rule.(uv)′=u′v+uv′.
(R5) Quotient rule.(vu)′=v2u′v−uv′, where v=0.
Proofs (sketch)
R4 proof: (u(x+h)v(x+h)−u(x)v(x))/h. Add and subtract u(x)v(x+h):
=h[u(x+h)−u(x)]v(x+h)+u(x)[v(x+h)−v(x)]=hu(x+h)−u(x)v(x+h)+u(x)hv(x+h)−v(x).
As h→0: u′(x)v(x)+u(x)v′(x). \qed
R5 proof: similar idea, with more bookkeeping. Or derive R5 from R4 plus (1/v)′=−v′/v2.
Standard derivatives table
f
f′
c (constant)
0
x
1
xn
nxn−1
x
1/(2x)
1/x
−1/x2
sinx
cosx
cosx
−sinx
tanx
sec2x
cotx
−csc2x
secx
secxtanx
cscx
−cscxcotx
ex
ex
ax
axlna
lnx
1/x
logax
1/(xlna)
Worked examples
Example 1.f(x)=5x3−4x+7. Find f′(x).
By R2, R3: f′(x)=15x2−4.
Example 2.f(x)=(3x+1)(2x−5). Find f′(x).
By R4: f′(x)=3⋅(2x−5)+(3x+1)⋅2=6x−15+6x+2=12x−13.
(Or expand first: f(x)=6x2−13x−5⇒f′(x)=12x−13. ✓)
Example 3.f(x)=x−1x2+1. Find f′(x).
By R5: f′=(x−1)22x⋅(x−1)−(x2+1)⋅1=(x−1)22x2−2x−x2−1=(x−1)2x2−2x−1.
Example 4.f(x)=xsinx. Find f′(x).
By R4: f′(x)=1⋅sinx+x⋅cosx=sinx+xcosx.
Example 5.f(x)=xsinx. Find f′(x).
By R5: f′(x)=x2cosx⋅x−sinx⋅1=x2xcosx−sinx.
Try it yourself
Find f′(x):
f(x)=7x4−3x2+2.
f(x)=x2+3sinx.
f(x)=(x2+1)(x−2).
f(x)=x+1x3.
f(x)=1−x1+x.
f(x)=(x+1)(x+2)(x+3). (Use R4 twice or expand.)
f(x)=x2cosx.
f(x)=1+cosxsinx.
f(x)=x⋅sinx.
f(x)=excosx.
f(x)=xlnx.
f(x)=1+x21.
Pitfalls / Tricks
Product rule: (uv)′=u′v′. Both terms must be present.
Quotient rule: numerator is u′v−uv′ (order matters).
For three-factor products, apply the product rule twice: (uvw)′=u′vw+uv′w+uvw′.
Insight. Derivatives are linear: (cu+dv)′=cu′+dv′. Whenever you see a linear combination, just differentiate term by term.