Introduction
A product can be difficult to differentiate when its factors have complicated forms. The logarithm offers a useful way to reorganize such expressions: it turns products into sums. The derivative of the logarithm then relates the rate of change of a positive function to the rate of change of the function itself.
The key identity is simple, but its conditions matter. The function must be positive wherever we take its natural logarithm, and it must be differentiable there. Under those assumptions, the derivative of its logarithm is its derivative divided by its value.
The quantity \(y'/y\) is called the logarithmic derivative of \(y\). It records the rate of change of \(y\) relative to its current value. We will derive the identity, establish how it applies to products and constant powers, and check each method in worked examples.
The identity and its conditions
Let \(I\) be an interval, and let \(y:I\to\mathbb{R}\) be differentiable and positive at every point of \(I\). Positivity means \(y(x)>0\) for each \(x\in I\), so \(\ln(y(x))\) is defined throughout the interval. The chain rule for the natural logarithm gives
This is a pointwise identity: at any \(x\in I\), both sides have the same value. The denominator is nonzero because \(y(x)>0\). A version written using \(y\) as a dependent variable is
When using this shorter notation, \(y\) still represents a differentiable function of \(x\), not a constant. The expression \(y'/y\) is not the derivative \(y'\) itself; it is the derivative divided by the function’s current value.
Why logarithms turn products into sums
For positive numbers \(u\) and \(v\), the logarithm law \(\ln(uv)=\ln u+\ln v\) converts multiplication into addition. Suppose \(u\) and \(v\) are positive differentiable functions on an interval. Their product is positive there, so we can apply the logarithmic derivative identity to it:
Differentiate both sides. The identity applies separately to each positive function, giving
Thus the logarithmic derivative of a product is the sum of the logarithmic derivatives of its factors. Equivalently, multiplying by \(uv\) recovers the ordinary product derivative:
The logarithmic form expresses the relative contributions of the two factors. The last equality also shows that it agrees with the product rule.
Why powers become multiples
Let \(u\) be positive and differentiable, and let \(r\) be a real constant. For positive \(u\), the real power \(u^r\) is defined by \(u^r=\exp(r\ln u)\), and it is positive. The logarithm law for powers gives \(\ln(u^r)=r\ln u\). Differentiating this equality and applying the identity to each positive function yields
So a constant power contributes a constant multiplier to the logarithmic derivative. Multiplying by \(u^r\) gives the equivalent ordinary derivative formula
The positivity condition is especially important when \(r\) is an arbitrary real number: the real-valued expression \(u^r=\exp(r\ln u)\) requires \(u>0\). For particular integer powers, differentiation may be defined more broadly, but the logarithmic argument still requires a positive base.
Worked examples
Worked Example: A product with an exponential factor
Let \(y=(x^2+1)e^{2x}\). Find \(y'/y\), then find \(y'\), and check both results at \(x=1\).
First check the domain condition. For every real \(x\), \(x^2+1>0\) and \(e^{2x}>0\), so \(y>0\). We can take the logarithm and use the product law:
Differentiate the two terms. By the chain rule, the derivative of \(\ln(x^2+1)\) is \(2x/(x^2+1)\), and the derivative of \(2x\) is \(2\). Therefore,
To recover \(y'\), multiply by \(y=(x^2+1)e^{2x}\):
At \(x=1\), the logarithmic derivative is \[ \frac{2}{2}+2=3. \] The function value is \(y(1)=2e^2\), so the identity gives \(y'(1)=3(2e^2)=6e^2\). Differentiating the original product directly gives \(y'=2xe^{2x}+2(x^2+1)e^{2x}\), which at \(x=1\) is \(2e^2+4e^2=6e^2\). The two calculations agree.
Worked Example: A positive function raised to a constant power
Let \(y=(2x+3)^4\), on the interval \(x>-3/2\). Find \(y'/y\) and \(y'\), then check the result at \(x=0\).
The condition \(x>-3/2\) ensures \(2x+3>0\), so \(y>0\) and \(\ln y\) is defined. Use the power law for logarithms:
Differentiate, using the chain rule for the logarithm of \(2x+3\):
Multiply by \(y\) to obtain the ordinary derivative:
At \(x=0\), the logarithmic derivative is \(8/3\), while \(y(0)=3^4=81\). Thus \(y'(0)=81(8/3)=216\). Differentiating the original expression by the chain rule gives \(4(2x+3)^3\cdot2=8(2x+3)^3\), which at \(x=0\) is \(8\cdot27=216\). This checks the result.
Worked Example: A product of positive powers
Let \(y=(x+1)^2(x+3)^3\) on \(x>-1\). Find its logarithmic derivative and evaluate \(y'\) at \(x=0\).
On this interval both \(x+1\) and \(x+3\) are positive. Therefore \(y>0\), and the logarithm can be expanded as a sum:
Differentiate each term:
At \(x=0\), this is \(2/1+3/3=2+1=3\). The function value is \(y(0)=1^2\cdot3^3=27\), so \(y'(0)=27\cdot3=81\).
Check directly by differentiating the product:
At \(x=0\), the first term is \(2\cdot1\cdot27=54\), and the second is \(3\cdot1\cdot9=27\). Their sum is \(54+27=81\), as obtained from the logarithmic derivative.
Relative rate of change
The identity also explains the interpretation of \(y'/y\). If \(y\) is positive, then \(y'\) describes its instantaneous rate of change, while dividing by \(y\) scales that rate by the function’s current size. For example, a function with a large value can have a large derivative simply because it is large; \(y'/y\) measures the rate relative to that value.
In differential notation, the identity can be written as
For a product of positive factors, each factor contributes its own relative rate, and those contributions add. For a positive function raised to a constant power, the relative rate is multiplied by that power. These conclusions follow from logarithm laws followed by differentiation, rather than from an additional, unrelated derivative rule.
A common mistake
A frequent error is to omit the division by the function and claim that \(\frac{d}{dx}\ln y=y'\). For \(y=(x^2+1)e^{2x}\) at \(x=1\), direct differentiation gives \(y'(1)=6e^2\). But the derivative of its logarithm is
not \(6e^2\). The correction is to apply the chain rule to the outer logarithm: its derivative contributes the factor \(1/y\), which must then be multiplied by the derivative \(y'\) of the inner function.
Another condition to check is positivity. If a function is not positive on the interval under consideration, the real expression \(\ln y\) may not be defined there. One cannot use the logarithmic derivative identity for that function on such an interval without first addressing the domain.
Summary
For a positive differentiable function, the logarithmic derivative identity is
It follows directly from the chain rule. Logarithm laws then show why products become sums and why positive constant powers become multiples:
In either case, multiply the logarithmic derivative by the original function to recover the ordinary derivative. Before applying the method, verify that the relevant functions are positive on the domain where the logarithm is used.
Check Your Understanding
Use the positivity conditions as well as the derivative calculations when answering.
- If \(y\) is positive and differentiable, what is \(\frac{d}{dx}\ln y\)?
- Why must a function be positive to apply the real logarithmic derivative identity as stated?
- If \(u\) and \(v\) are positive differentiable functions, write the logarithmic derivative of \(uv\) as a sum.
- For positive differentiable \(u\) and constant \(r\), write the logarithmic derivative of \(u^r\).
- If \(y(2)=5\) and \(y'(2)=15\), what is \(\left.\frac{d}{dx}\ln y\right|_{x=2}\)?
