Differentiation: (lesson 2 of 3)
Product and Quotient Rule for Derivatives
Product Rule
Let f(x) and g(x) be differentiable at x.
Then h(x)=f(x)g(x) is differentiable at x and
(f(x)⋅g(x))′=f′(x)⋅g(x)+f(x)⋅g′(x)
We illustrate product rule with the following examples:
Example 1:
(x3⋅ex)′=(x3)′⋅(ex)′+x3⋅(ex)′=3⋅x2⋅ex+x3⋅ex==x2ex(3+x)
Example 2:
((x+2)⋅(x2−x+1))′=(x+2)′⋅(x2−x+1)+(x+2)⋅(x2−x+1)′==1⋅(x2−x+1)+(x+2)⋅(2x−1)==x2−x+1+x2−x+4x−2==2x2+2x−1
Quotient rule:
Let f(x) and g(x) be differentiable at x with
g(x)=0. Then g(x)f(x) is differentiable at x and
[g(x)f(x)]′=[g(x)]2f′(x)⋅g(x)−f(x)⋅g′(x)
We illustrate quotient rule with the following examples:
Example 3:Differentiate
y=2+xx2
Solution 3:
(2+xx2)′=(2+x)2(x2)′⋅(2+x)−x2⋅(2+x)′==(2+x)22x⋅(2+x)−x2⋅(0+1)==(2+x)24x+2x2−x2==(2+x)2x2+4x
Example 4:Differentiate
y=1+x2lnx
Solution 4:
(x2lnx)′=(x2)2(lnx)′⋅x2−lnx⋅(x2)==x4x1⋅x2−lnx⋅2x==x4x−2xlnx==x4x(1−2lnx)==x31−2lnx
Example 5:Differentiate y=tanx
Solution 5:
(cosxsinx)′=(cosx)2(sinx)′⋅cosx−sinx⋅(cosx)′==cos2xcosx⋅cosx−sinx⋅(−sinx)==cos2xcos2x+sin2x==cos2x1
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