> DIFFERENTIATION

Created by Joanne Lee|

Differentiation Rules:

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Basic Differentiation Rules

  • 1. Derivative of a constant function
    > The derivative of a constant function is 0.
    > ex. if f(t) = -34, f'(t) = 0

  • 2. The Single Variable Rule
    > The derivative of x is 1.
    > ex. if f(x) = x, f'(x) = 1
    > ex. if s(t) = t, s'(t) = 1

  • 3. The Power Rule
    > If n is a rational number, then the function xn is differentiable and (d/dx)[xn] = n * xn-1.
    > ex. if f(t) = x4, f'(t) = 4x3

  • 4. The General Power Rule
    > The derivative of the term axn, where a and n are real numbers, is
    (a * n)xn-1.
    > ex. if g(x) = -5t3, g'(x) = -15t2

  • 5. The Sum & Difference Rules
    > The derivative of a sum or difference is the sum or difference of the derivatives.
    > (d/dx)[f(x) + g(x)] = f'(x) + g'(x)
    > (d/dx)[f(x) - g(x)] = f'(x) - g'(x)

The Product and Quotient Rules

  • The Product Rule: If a function is the product of two differential functions (ex. f(x) and g(x)), the derivative is

    f'(x) * g(x) + f(x) * g'(x)

  • The Quotient Rule: If a function is the quotient of two differential functions (ex. f(x) and g(x)), the derivative is

    f'(x) * g(x) - f(x) * g'(x)
    [g(x)]2

The Chain Rule

If f and g are both differentiable, then f(g(x)) is also differentiable and
f'(g(x)) * g'(x).
Example: Find f'(x) if f(x) = (5x+3)2
  • 1. f'(x) = 2(5x+3) * (5x+3)'
  • 2. f'(x) = 2(5x+3)(5)
  • 3. f'(x) = 10(5x+3)

Implicit Differentiation

Basic facts to consider: In implicit differentiation, you will have a (dy/dx) for each y in the original function or equation. Isolate the (dy/dx). If you are taking the second derivative, you will often substitute the expression you found for the first derivative somewhere in the process.
  • > Differentiation is taking place with respect to x.
  • > When differentiating terms involving x alone, differentiate as usual.
  • > When differentiating terms involving y, apply the chain rule (it is assumed that y is defined implicitly as differentiable function of x.
How to solve an implicit differentiation problem:
  • 1. Differentiate both sides with respect to x. Multiply by dy/dx every time you differentiate an expression containing y (apply chain rule).
  • 2. Isolate dy/dx by performing the necessary steps to transfer all of the non-dy/dx terms onto one side of the equation with the dy/dx on the other.
  • 3. Factor out dy/dx if necessary.
  • 4. Solve for dy/dx and find the answer.

Derivatives of Inverse Functions

A function g is the inverse function of function f if f(g(x)) = x for each x in the domain of g and g(f(x)) = x for each x in the domain of f. The function g(x) is denoted by f-1(x).

To find the derivative of an inverse function: Let f and g be inverse functions, such that f(g(x)) = x = g(f(x)) where f(a) = b and
g(b) = a.

Finding g'(b) for a point (a, b) on f'(x):
  • 1. Find the value of f'(x)
  • 2. If you are only given b, set b = f(x) to find a.
  • 3. Find f'(a)
  • 4. g'(b) = 1/(f'(a))

  • > Essentially: (f-1)'(x) =    1   
                              f'(f-1(x))

Derivatives List

Basic Derivatives:
  • > (sinx)' = cosx
  • > (cosx)' = -sinx
  • > (tanx)' = sec2x
  • > (cotx)' = -csc2x
  • > (secx)' = secxtanx
  • > (cscx)' = -cscxcotx
  • > (lnu)' = 1/u(du/dx)
  • > (eu)' = eu(du/dx)

  • * Where u is a function of x, and a is a constant
  Derivatives of Trig Functions:
  • > (sin-1x)' =   1  
                 √(1-x2)
  • > (cos-1x)' =     1  
                   √(1-x2)
  • > (tan-1x)' =   1  
                 (1+x2)
  • > (cot-1x)' =     1  
                   (1+x2)
  • > (sec-1x)' =       1    
                   |x|*√(x2-1)
  • > (csc-1x)' =         1    
                     |x|*√(x2-1)