Differentiation of Logarithmic Functions

Examples of the derivatives of logarithmic functions, in calculus, are presented. Several examples, with detailed solutions, involving products, sums and quotients of exponential functions are examined.

The derivative of f(x) = log b x is given by

f '(x) = 1 / (x ln b)


Note: if f(x) = ln x , then f '(x) = 1 / x

Example 1: Find the derivative of f(x) = log 3 x

Solution to Example 1:

  • Apply the formula above to obtain

    f '(x) = 1 / (x ln 3)


Example 2: Find the derivative of f(x) = ln x + 6x 2

Solution to Example 2:

  • Let g(x) = ln x and h(x) = 6x 2, function f is the sum of functions g and h: f(x) = g(x) + h(x). Use the sum rule, f '(x) = g '(x) + h '(x), to find the derivative of function f

    f '(x) = 1 / x + 12x


Example 3: Find the derivative of f(x) = log 3 x / ( 1 - x )

Solution to Example 3:

  • Let g(x) = log 3 x and h(x) = 1 - x, function f is the quotient of functions g and h: f(x) = g(x) / h(x). Hence we use the quotient rule, f '(x) = [ h(x) g '(x) - g(x) h '(x) ] / h(x) 2, to find the derivative of function f.

    g '(x) = 1 / (x ln 3)

    h '(x) = -1

    f '(x) = [ h(x) g '(x) - g(x) h '(x) ] / h(x) 2

    = [ (1 - x)(1 / (x ln 3)) - (log 3 x)(-1) ] / (1 - x) 2



Example 4: Find the derivative of f(x) = ln (-4x + 1)

Solution to Example 4:

  • Let u = -4x + 1 and y = ln u, Use the chain rule to find the derivative of function f as follows.

    f '(x) = (dy / du) (du / dx)

  • dy / du = 1 / u and du / dx = -4

    f '(x) = (1 / u)(-4) = -4 / u

  • Substitute u = -4x + 1 in f '(x) above

    f '(x) = -4 / (-4x + 1)
Exercises Find the derivative of each function.

1 - f(x) = ln(x
2)

2 - g(x) = ln x - x
7

3 - h(x) = ln x / (2x - 3)

4 - j(x) = ln (x + 3) ln (x - 1)

solutions to the above exercises

1 - f '(x) = 2 / x

2 - g '(x) = 1 / x -7x
6

3 - h '(x) = (2x - 3 - 2x ln x) / [ x(2x -3)
2 ]

4 - j '(x) = ln (x + 3) / (x - 1) + ln (x - 1) / (x + 3)

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differentiation and derivatives