Econ 300: Calculus Review


Some Basic Rules



Rule. Let r be a fixed number. Assume that f(x)=xr. Then f '(x)=rxr-1.

Some Applications
  • f(x)=x3. Then r=3. Thus f '(x)=3x2.
  • f(x)=x1/3. Then r=1/3. Thus f '(x)=(1/3)x-2/3.
    Recall that xr=1/(x-r).
    Thus we can also write the derivative as f '(x)=1/(3x2/3).
  • f(x)=a, where a is a fixed number. Then f(x)=ax0. Thus, f '(x)=0.
  • f(x)=1/x. This can be written as f(x)=x-1. Thus r=-1.
    Consequently f '(x)=-x-2 which can be written as f '(x)=-1/(x2).
Rule. Let f(x)=ln (x). Then f '(x)=1/x.




Linearity



Rule. Assume a function f can be written as the sum of two other functions, i.e., f(x)=g(x)+h(x).
Then f '(x)=g '(x)+h '(x).

Some Applications
  • f(x)=x3+a, where a is a fixed number. Then g(x)=x3 and h(x)=a.
    From above it follows that g '(x)=3x2 and h '(x)=0.
    Thus f '(x)=g '(x)+h '(x)=3x2+0= 3x2.
  • f(x)=x+(1/x). Then g(x)=x and h(x)=1/x.
    Then g '(x)=1 and h '(x)=-1/(x2).
    Thus f '(x)=g '(x)+h '(x)=1-1/(x2).

Rule. Assume a function f can be written as the product of fixed number a and some other function, i.e., f(x)=ag(x).
Then f '(x)=ag '(x).

Some Applications
  • f(x)=10x then f (x)=10g(x) where g(x)=x.
    Because g '(x)=1 we get f '(x)=10.
  • f(x)=5x2 then f (x)=5g(x) where g(x)=x2.
    Thus, f '(x)=10x because the derivative of g(x) is 2x.


Product Rule



Rule. Assume a function f can be written as the product of two other functions, i.e., f(x)=g(x)h(x).
Then f '(x)=g '(x)h(x)+g(x)h'(x).

Some Applications
  • f(x)=xln (x). Then g(x)=x and h(x)=ln (x).
    Thus, g '(x)=1 and h '(x)=1/x.
    Consequently f '(x)=g '(x)h(x)+g(x)h '(x)=ln (x)+x(1/x)=ln (x)+1.
  • f(x)=(1+x)/x2. This can be written as f(x)=(1+x)x-2. Thus, g(x)=1+x and h(x)=x-2.
    Therefore, applying additivity we get g '(x)=1. Moreover, applying the first rule explained in this review, we get h'(x)=-2x-3.
    Thus f '(x)=g '(x)h(x)+g(x)h'(x)=x-2+(1+x)(-2x-3).
    This can be rewritten as f '(x)=1/(x2)-2(1+x)/x3.


Chain Rule



Rule. Assume a function f can be written as f(x)=g(h(x)).
Then f '(x)=g '(h(x))h'(x).

Some Applications
  • f(x)=(1+4x)3. Then h(x)=1+4x. Moreover, g(x)=x3.
    Note that if we insert in g(x) for x the expression h(x)=1+4x then we get f(x).
    We now want to find the derivative of f(x). Using the chain rule, we first take the derivative of g(x)=x3.
    This derivative is g '(x)=3x2.
    Now insert for x again h(x)=1+x. Thus, g'(h(x))=3(1+4x)2.
    Finally, we must multiply this with h '(x)=4.
    Consequently, f '(x)=12(1+4x)2.
  • f(x)=(1+x2)1/3. Then h(x)=1+x2 and g(x)=x1/3.
    As shown above, the derivative of g(x)=x1/3 is g '(x)=(1/3)x-2/3.
    Now insert h(x)=1+x2 for x in g '(x).
    Then g'(h(x))=(1/3)(1+x2)-2/3.
    Because of additivity h '(x)=2x.
    Thus, f '(x)=(2/3)x(1+x2)-2/3.


Quotient Rule



Rule. Assume a function f can be written as the quotient of two other functions, i.e., f(x)=g(x)/h(x).
Then

f '(x)= g '(x)h(x)-g(x)h'(x)

h(x)2

 

Some Applications

  • f(x)=x/(1+x3). Then g(x)=x and h(x)=1+x3, and therefore g'(x)=1 and h '(x)=3x2 (using additivity). Thus

      f'(x)=[1+x3-3x2]/ (1+x3)2.

  • f(x)=x3/(1+x2). Then g(x)=x3 and h(x)=1+x2, and therefore g '(x)=3x2 and h '(x)=2x. Thus

    f'(x)=[ 3x2(1+x2)-x32x ] / (1+x2)2 = [ 3x2+x4 ] / (1+x2)2 .

Remark: The quotient rule follows from the product rule and the chain rule. In particular f(x) can be written as f(x)=g(x)h(x)-1.
Then f '(x)=g '(x)h(x)-1+ g(x)(h(x)-1)', by the product rule. Now note that the chain rule implies
(h(x)-1)' =(-1)h(x)-2h '(x)=-h(x)/ h(x)2.
Thus, f '(x)=g '(x)h(x)-1-g(x)h(x)/ h(x)2.
Using the fact that h(x)-1=1/h(x)=h(x)/h(x)2, you immediately get the formula for the quotient rule.

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