Product Rule
If f and g are differentiable, then:
dxd[f(x)⋅g(x)]=f(x)⋅g′(x)+g(x)⋅f′(x)
Memory Aid
"First times derivative of second, plus second times derivative of first"
Or using Leibniz notation:
dxd[uv]=udxdv+vdxdu
Quotient Rule
If f and g are differentiable and g(x)=0:
dxd[g(x)f(x)]=[g(x)]2g(x)⋅f′(x)−f(x)⋅g′(x)
Memory Aid
"Low d-high minus high d-low, over low squared"
Or using Leibniz notation:
dxd[vu]=v2vdxdu−udxdv
Extended Product Rule
For three functions:
dxd[f⋅g⋅h]=f′⋅g⋅h+f⋅g′⋅h+f⋅g⋅h′
Examples
Example 1: Product Rule
Find the derivative of f(x)=(x2+1)(x3−2x):
f′(x)=(x2+1)⋅(3x2−2)+(x3−2x)⋅(2x)
=3x4−2x2+3x2−2+2x4−4x2
=5x4−3x2−2
Example 2: Quotient Rule
Find the derivative of f(x)=x−3x2+1:
f′(x)=(x−3)2(x−3)(2x)−(x2+1)(1)
=(x−3)22x2−6x−x2−1
=(x−3)2x2−6x−1
Example 3: Product with trig
Find dxd[x2sin(x)]:
=x2⋅cos(x)+sin(x)⋅2x=x2cos(x)+2xsin(x)
Example 4: Quotient with exponential
Find dxd[x+1ex]:
=(x+1)2(x+1)⋅ex−ex⋅1
=(x+1)2ex(x+1−1)
=(x+1)2xex
Tips
- Always simplify before differentiating if possible
- For quotients like f(x)1, you can use: dxd[f1]=−f2f′
- Sometimes it's easier to rewrite as a product: gf=f⋅g−1