Logarithmic Differentiation, calculus lesson

Differentiating a product of five factors with the product rule produces five long terms and plenty of room for mistakes. Logarithms turn products into sums and powers into multiples, and sums are easy to differentiate. So it often pays to take the logarithm first, differentiate, and then undo the logarithm at the end.

Key terms

Logarithmic differentiation

Definition Let $y = f(x)$ be differentiable at $x$ with $f(x) \neq 0$ . Then

\frac{d}{dx} \ln \lvert f(x) \rvert = \frac{f'(x)}{f(x)}, \qquad \text{so} \qquad f'(x) = f(x) \cdot \frac{d}{dx} \ln \lvert f(x) \rvert.

Logarithmic differentiation means taking $\ln \lvert y \rvert$ , simplifying with the laws of logarithms, differentiating with respect to $x$ , and multiplying the result by $y$ .

In plain words The quantity $\frac{y'}{y}$ is the relative rate of change of $y$ . It is often much easier to compute than $y'$ itself, and you recover $y'$ by multiplying by $y$ .

Example For $y = x^5 e^{3x}$ with $x \neq 0$ , taking logarithms gives $\ln \lvert y \rvert = 5 \ln \lvert x \rvert + 3x$ . Differentiating gives $\frac{y'}{y} = \frac{5}{x} + 3$ , so

y' = x^5 e^{3x}\left(\frac{5}{x} + 3\right) = (5x^4 + 3x^5)e^{3x}.

The final formula also holds at $x = 0$ , as the product rule confirms.

Products

Definition If $y = f_1(x) f_2(x) \cdots f_n(x)$ where each $f_i$ is differentiable at $x$ and $f_i(x) \neq 0$ , then

\frac{y'}{y} = \frac{f_1'(x)}{f_1(x)} + \frac{f_2'(x)}{f_2(x)} + \cdots + \frac{f_n'(x)}{f_n(x)}.

In plain words The relative rate of change of a product is the sum of the relative rates of change of its factors.

Example Let $y = x(x + 1)(x + 2)$ at $x = 1$ . Here $y = 6$ and

\frac{y'}{y} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{11}{6},

so $y' = 6 \cdot \frac{11}{6} = 11$ . Expanding gives $y = x^3 + 3x^2 + 2x$ and $y' = 3x^2 + 6x + 2$ , which is also $11$ at $x = 1$ .

Quotients

Definition Let $f$ and $g$ be differentiable at $x$ with $f(x) \neq 0$ and $g(x) \neq 0$ , and let $y = f/g$ . Then near $x$

\ln \lvert y \rvert = \ln \lvert f \rvert - \ln \lvert g \rvert,

and at $x$

\frac{y'}{y} = \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)}.

Factors in the denominator contribute with a minus sign, and powers come down as coefficients.

In plain words Division becomes subtraction. A power like $(x^2 + 1)^2$ in the denominator just becomes $-2 \ln(x^2 + 1)$ .

Example Take

y = \frac{(x + 1)^3}{(x^2 + 1)^2}

at $x = 1$ , where $y = \frac{8}{4} = 2$ . Taking logarithms gives

\ln \lvert y \rvert = 3 \ln \lvert x + 1 \rvert - 2 \ln(x^2 + 1),

and differentiating gives

\frac{y'}{y} = \frac{3}{x + 1} - \frac{4x}{x^2 + 1} = \frac{3}{2} - 2 = -\frac{1}{2},

so $y'(1) = 2 \cdot \left(-\frac{1}{2}\right) = -1$ .

Absolute values for nonzero real factors

Definition The real logarithm is defined only for positive numbers, so for a function that may be negative one works with $\ln \lvert y \rvert$ . The rules

\ln \lvert uv \rvert = \ln \lvert u \rvert + \ln \lvert v \rvert, \qquad \ln \left\lvert \frac{u}{v} \right\rvert = \ln \lvert u \rvert - \ln \lvert v \rvert, \qquad \ln \lvert u^n \rvert = n \ln \lvert u \rvert

hold whenever $u$ and $v$ are nonzero real numbers and $n$ is an integer. For a function $u$ differentiable at $x$ with $u(x) \neq 0$ , the derivative of $\ln \lvert u \rvert$ at $x$ is $\frac{u'}{u}$ . The resulting formula for $y'$ is proved only at points where every factor is nonzero. Points where a factor is zero must be checked separately.

In plain words The absolute value bars let you use the method even where $y$ is negative, and they disappear from the answer because the derivative of $\ln \lvert u \rvert$ is $u'/u$ , with no bars.

Example Let $y = \frac{x - 3}{x + 1}$ at $x = 0$ , where $y = -3 < 0$ , so $\ln y$ is undefined there. Use the absolute value instead.

\ln \lvert y \rvert = \ln \lvert x - 3 \rvert - \ln \lvert x + 1 \rvert

Then

\frac{y'}{y} = \frac{1}{x - 3} - \frac{1}{x + 1} = -\frac{1}{3} - 1 = -\frac{4}{3},

so $y'(0) = -3 \cdot \left(-\frac{4}{3}\right) = 4$ . The quotient rule agrees, since $y' = \frac{4}{(x + 1)^2}$ .

Variable exponents

Definition Let $f$ and $g$ be differentiable at $x$ with $f(x) > 0$ . Wherever $f > 0$ the function $y = f(x)^{g(x)}$ is defined as $e^{g(x) \ln f(x)}$ , and $f$ stays positive near $x$ because it is continuous there. Taking logarithms gives $\ln y = g(x) \ln f(x)$ , and differentiating gives

y' = f(x)^{g(x)} \left( g'(x) \ln f(x) + g(x) \frac{f'(x)}{f(x)} \right).

In plain words When both the base and the exponent change with $x$ , neither the power rule (fixed exponent) nor the exponential rule (fixed base) applies. Taking a logarithm brings the exponent down where the product rule can handle it.

Example For $y = x^x$ with $x > 0$ , taking logarithms gives $\ln y = x \ln x$ , so $\frac{y'}{y} = \ln x + 1$ and

y' = x^x(\ln x + 1).

At $x = 2$ this is $4(\ln 2 + 1) \approx 6.773$ .

Positive base assumptions

Definition When the exponent varies with $x$ , the power $f(x)^{g(x)}$ is defined as $e^{g(x) \ln f(x)}$ , which requires $f(x) > 0$ . The formula for variable exponents therefore applies only on intervals where the base is positive. A negative base has a real power only for some exponents, namely integers and fractions in lowest terms with an odd denominator. For example $(-8)^{1/3} = -2$ , while $(-8)^{1/2}$ is not real. Suppose the base is negative at a point and the exponent $g$ is continuous and not constant on any open interval around that point. Then every such interval contains points where $g$ is irrational and the base is still negative, so the power is undefined at points arbitrarily close to the point and cannot be differentiated there. The formula with $\ln f(x)$ makes no sense there either.

In plain words Check that the base is positive before you use the formula. If it is not, restrict to an interval where it is.

Example The base of $y = (x^2 + 1)^x$ is always positive, so the formula holds on all of $\mathbb{R}$ . With $\ln y = x \ln(x^2 + 1)$ ,

y' = (x^2 + 1)^x \left( \ln(x^2 + 1) + \frac{2x^2}{x^2 + 1} \right).

At $x = 1$ this is $2(\ln 2 + 1) \approx 3.386$ . In contrast, $x^x$ cannot be differentiated at any $x < 0$ . It has real values at some negative points, such as $(-1)^{-1} = -1$ , but not at others arbitrarily close to them, such as $x = -\frac{1}{2}$ , where $\left(-\frac{1}{2}\right)^{-1/2}$ is not a real number.

Check yourself

Differentiate $y = x^{\sin x}$ for $x > 0$ .

Show the answer

Taking logarithms gives $\ln y = \sin x \ln x$ , so $\frac{y'}{y} = \cos x \ln x + \frac{\sin x}{x}$ and

y' = x^{\sin x}\left(\cos x \ln x + \frac{\sin x}{x}\right).

Let $y = x^2 (x - 1)^3 (x + 2)$ . Use logarithmic differentiation to find $y'(2)$ .

Show the answer

At $x = 2$ the value is $y = 4 \cdot 1 \cdot 4 = 16$ , and the relative rate is

\frac{y'}{y} = \frac{2}{x} + \frac{3}{x - 1} + \frac{1}{x + 2} = 1 + 3 + \frac{1}{4} = \frac{17}{4}.

So $y'(2) = 16 \cdot \frac{17}{4} = 68$ .

Why do we write $\ln \lvert y \rvert$ rather than $\ln y$ for $y = \frac{x - 3}{x + 1}$ near $x = 0$ ?

Show the answer

Near $x = 0$ the value of $y$ is negative (it is $-3$ at $x = 0$ ), and $\ln$ of a negative number is undefined. The absolute value makes the logarithm defined, and the derivative of $\ln \lvert u \rvert$ is still $\frac{u'}{u}$ .

Logarithmic Differentiation

Theory, part 61

Practice, part 62

Key terms