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Proof

If x > 0 and y = ln x , then e y = x . Differentiating both sides of this equation results in the equation

e y d y d x = 1 .

Solving for d y d x yields

d y d x = 1 e y .

Finally, we substitute x = e y to obtain

d y d x = 1 x .

We may also derive this result by applying the inverse function theorem, as follows. Since y = g ( x ) = ln x is the inverse of f ( x ) = e x , by applying the inverse function theorem we have

d y d x = 1 f ( g ( x ) ) = 1 e ln x = 1 x .

Using this result and applying the chain rule to h ( x ) = ln ( g ( x ) ) yields

h ( x ) = 1 g ( x ) g ( x ) .

The graph of y = ln x and its derivative d y d x = 1 x are shown in [link] .

Graph of the function ln x along with its derivative 1/x. The function ln x is increasing on (0, + ∞). Its derivative is decreasing but greater than 0 on (0, + ∞).
The function y = ln x is increasing on ( 0 , + ) . Its derivative y = 1 x is greater than zero on ( 0 , + ) .

Taking a derivative of a natural logarithm

Find the derivative of f ( x ) = ln ( x 3 + 3 x 4 ) .

Use [link] directly.

f ( x ) = 1 x 3 + 3 x 4 · ( 3 x 2 + 3 ) Use g ( x ) = x 3 + 3 x 4 in h ( x ) = 1 g ( x ) g ( x ) . = 3 x 2 + 3 x 3 + 3 x 4 Rewrite.
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Using properties of logarithms in a derivative

Find the derivative of f ( x ) = ln ( x 2 sin x 2 x + 1 ) .

At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms prior to finding the derivative, we can make the problem much simpler.

f ( x ) = ln ( x 2 sin x 2 x + 1 ) = 2 ln x + ln ( sin x ) ln ( 2 x + 1 ) Apply properties of logarithms. f ( x ) = 2 x + cot x 2 2 x + 1 Apply sum rule and h ( x ) = 1 g ( x ) g ( x ) .
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Differentiate: f ( x ) = ln ( 3 x + 2 ) 5 .

f ( x ) = 15 3 x + 2

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Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of y = l o g b x and y = b x for b > 0 , b 1 .

Derivatives of general exponential and logarithmic functions

Let b > 0 , b 1 , and let g ( x ) be a differentiable function.

  1. If, y = log b x , then
    d y d x = 1 x ln b .

    More generally, if h ( x ) = log b ( g ( x ) ) , then for all values of x for which g ( x ) > 0 ,
    h ( x ) = g ( x ) g ( x ) ln b .
  2. If y = b x , then
    d y d x = b x ln b .

    More generally, if h ( x ) = b g ( x ) , then
    h ( x ) = b g ( x ) g ( x ) ln b .

Proof

If y = log b x , then b y = x . It follows that ln ( b y ) = ln x . Thus y ln b = ln x . Solving for y , we have y = ln x ln b . Differentiating and keeping in mind that ln b is a constant, we see that

d y d x = 1 x ln b .

The derivative in [link] now follows from the chain rule.

If y = b x , then ln y = x ln b . Using implicit differentiation, again keeping in mind that ln b is constant, it follows that 1 y d y d x = ln b . Solving for d y d x and substituting y = b x , we see that

d y d x = y ln b = b x ln b .

The more general derivative ( [link] ) follows from the chain rule.

Applying derivative formulas

Find the derivative of h ( x ) = 3 x 3 x + 2 .

Use the quotient rule and [link] .

h ( x ) = 3 x ln 3 ( 3 x + 2 ) 3 x ln 3 ( 3 x ) ( 3 x + 2 ) 2 Apply the quotient rule. = 2 · 3 x ln 3 ( 3 x + 2 ) 2 Simplify.
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Finding the slope of a tangent line

Find the slope of the line tangent to the graph of y = log 2 ( 3 x + 1 ) at x = 1 .

To find the slope, we must evaluate d y d x at x = 1 . Using [link] , we see that

d y d x = 3 ln 2 ( 3 x + 1 ) .

By evaluating the derivative at x = 1 , we see that the tangent line has slope

d y d x | x = 1 = 3 4 ln 2 = 3 ln 16 .
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Find the slope for the line tangent to y = 3 x at x = 2 .

9 ln ( 3 )

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Logarithmic differentiation

At this point, we can take derivatives of functions of the form y = ( g ( x ) ) n for certain values of n , as well as functions of the form y = b g ( x ) , where b > 0 and b 1 . Unfortunately, we still do not know the derivatives of functions such as y = x x or y = x π . These functions require a technique called logarithmic differentiation    , which allows us to differentiate any function of the form h ( x ) = g ( x ) f ( x ) . It can also be used to convert a very complex differentiation problem into a simpler one, such as finding the derivative of y = x 2 x + 1 e x sin 3 x . We outline this technique in the following problem-solving strategy.

Questions & Answers

Find the derivative of g(x)=−3.
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f(x) = x-2 g(x) = 3x + 5 fog(x)? f(x)/g(x)
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fog(x)= f(g(x)) = x-2 = 3x+5-2 = 3x+3 f(x)/g(x)= x-2/3x+5
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(mathematics) For a complex number a+bi, the principal square root of the sum of the squares of its real and imaginary parts, √a2+b2 . Denoted by | |. The absolute value |x| of a real number x is √x2 , which is equal to x if x is non-negative, and −x if x is negative.
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Practice Key Terms 1

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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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