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Copyright © Cengage Learning. All rights reserved. 3 Differentiation Rules Copyright © Cengage Learning. All rights reserved. 1

3.6 Derivatives of Logarithmic Functions Copyright © Cengage Learning. All rights reserved. 2

Derivatives of Logarithmic Functions In this section we use implicit differentiation to find the derivatives of the logarithmic functions y = logb x and, in particular, the natural logarithmic function y = ln x. [It can be proved that logarithmic functions are differentiable; this is certainly plausible from their graphs (see Figure 12 in Section 1.5).] Figure 12

Derivatives of Logarithmic Functions In general, if we combine Formula 2 with the Chain Rule, we get or

Example 2 Find ln(sin x). Solution: Using (3), we have

Logarithmic Differentiation

Logarithmic Differentiation The calculation of derivatives of complicated functions involving products, quotients, or powers can often be simplified by taking logarithms. The method used in the next example is called logarithmic differentiation.

Example 7 Differentiate Solution: We take logarithms of both sides of the equation and use the Laws of Logarithms to simplify: ln y = ln x + ln(x2 + 1) – 5 ln(3x + 2) Differentiating implicitly with respect to x gives

Example 7 – Solution Solving for dy/dx, we get cont’d Solving for dy/dx, we get Because we have an explicit expression for y, we can substitute and write

Logarithmic Differentiation

The Number e as a Limit

The Number e as a Limit If f (x) = ln x, then f (x) = 1/x. Thus f (1) = 1. We now use this fact to express the number e as a limit. From the definition of a derivative as a limit, we have

The Number e as a Limit Because f (1) = 1, we have Then, by the continuity of the exponential function, we have

The Number e as a Limit Formula 5 is illustrated by the graph of the function y = (1 + x)1/x in Figure 4 and a table of values for small values of x. This illustrates the fact that, correct to seven decimal places, e  2.7182818 Figure 4

The Number e as a Limit If we put n = 1/x in Formula 5, then n  as x  0+ and so an alternative expression for e is