Why Compare a Derivative with the Function Value?
A derivative \(f'(x)\) measures the instantaneous change in \(f\) per unit change in \(x\). Its numerical size, however, depends on the scale of the function. A change of 10 units is substantial for a quantity near 5 but small for one near 10,000.
Dividing the derivative by the current function value gives a rate relative to the function's size. This is especially useful for variable-power functions: logarithmic differentiation often finds this quotient more directly than it finds the derivative itself.
The phrase “relative rate” emphasizes comparison with the current value of the function. If \(x\) represents time, then \(f'(x)/f(x)\) is a relative rate per unit time. If \(x\) is a different kind of input, it is a relative rate per unit of that input. The quotient does not automatically represent a percentage: to express it as a percentage rate, multiply its numerical value by \(100\%\).
Relative Rate as Fractional Change
The definition follows directly from the derivative. For a small input change \(h\), the change in the function is approximately \(f'(x)h\). Dividing that change by the starting value \(f(x)\) gives an approximate fractional change:
More precisely, because \(f(x)\ne 0\), the definition of derivative gives
Thus the relative rate is the instantaneous rate of change of the function divided by its present value. For small \(h\), multiplying the relative rate by \(h\) estimates the fractional change \( [f(x+h)-f(x)]/f(x)\).
For positive-valued functions, there is another useful interpretation. The chain rule gives
This identity explains why logarithmic differentiation naturally produces relative rates. It also sets a condition: the real logarithm \(\ln f(x)\) is defined only when \(f(x)>0\), even though the quotient \(f'(x)/f(x)\) is defined whenever \(f(x)\ne 0\).
Relative Rates for Variable-Power Functions
Consider \(y=u(x)^{v(x)}\), where \(u\) and \(v\) are differentiable and \(u(x)>0\) on the interval being considered. The positive-base condition ensures that the real-valued logarithm \(\ln u(x)\) is defined. Since \(y>0\), take logarithms and differentiate:
The product rule is used on the right-hand side, and the chain rule gives \(d(\ln u)/dx=u'/u\). The resulting expression is the relative rate. Multiplying by \(y\) recovers the ordinary derivative.
The formula has two contributions. The term \(v'(x)\ln u(x)\) accounts for the changing exponent. The term \(v(x)u'(x)/u(x)\) accounts for the changing base, measured relative to the base itself. If both the base and exponent vary, both terms are needed.
Worked Examples
Worked Example: The Relative Rate of \(x^x\)
Let \(y=x^x\), with \(x>0\). Find its relative rate and derivative at \(x=2\).
Here \(u(x)=x\), \(v(x)=x\), \(u'(x)=1\), and \(v'(x)=1\). The variable-power formula gives
At \(x=2\), the relative rate is \(\ln 2+1\). Also \(y(2)=2^2=4\), so the derivative is
To check, write \(x^x=\exp(x\ln x)\). Differentiating directly by the chain rule gives \[ y'=\exp(x\ln x)\left(\ln x+1\right)=x^x(\ln x+1). \] At \(x=2\), this is \(4(\ln2+1)\), agreeing with the result. The relative rate is \(\ln2+1\), not the derivative itself; the derivative includes the factor \(y(2)=4\).
Worked Example: A Shifted Base, \((x+1)^{2x}\)
Let \(y=(x+1)^{2x}\), for \(x>-1\), so the base is positive. Find the relative rate and derivative at \(x=1\).
Take \(u(x)=x+1\) and \(v(x)=2x\). Then \(u'(x)=1\) and \(v'(x)=2\), so
At \(x=1\), the relative rate is
The function value is \(y(1)=2^2=4\), hence
For a check, express \(y=\exp(2x\ln(x+1))\). Differentiating the exponent gives \(2\ln(x+1)+2x/(x+1)\), so the chain rule gives \[ y'=\exp(2x\ln(x+1)) \left(2\ln(x+1)+\frac{2x}{x+1}\right). \] At \(x=1\), the exponential factor is \(\exp(2\ln2)=4\), giving \(4(2\ln2+1)\) as before.
Worked Example: A Quadratic Base and a Linear Exponent
Let \(y=(x^2+1)^{3x}\). Find its relative rate and derivative at \(x=1\).
The base \(u(x)=x^2+1\) is positive for every real \(x\), and \(u'(x)=2x\). The exponent is \(v(x)=3x\), with \(v'(x)=3\). Therefore
At \(x=1\), \(y(1)=2^3=8\), and the relative rate is
Thus
To check, write \(y=\exp(3x\ln(x^2+1))\). Differentiating the exponent yields \(3\ln(x^2+1)+3x(2x)/(x^2+1)\), and the chain rule multiplies this by \(y\). Substitution of \(x=1\) gives \(8(3\ln2+3)\), as required.
What the Sign of a Relative Rate Says
The sign of \(f'(x)/f(x)\) must be interpreted with the sign of \(f(x)\) in mind. If \(f(x)>0\), then \(f'(x)/f(x)\) has the same sign as \(f'(x)\): a positive relative rate indicates that \(f\) is increasing at that point, and a negative relative rate indicates that it is decreasing. If \(f(x)<0\), the quotient has the opposite sign from \(f'(x)\).
For example, let \(f(x)=-e^x\). Then \(f'(x)=-e^x\), so
The function is decreasing, even though its relative rate is positive. In this case, the positive relative rate indicates that the magnitude \(|f(x)|=e^x\) is increasing. More generally, wherever \(f\ne0\), \(f'/f\) is the derivative of \(\ln|f|\). This follows from the chain rule applied to \(\ln|f(x)|\). The logarithmic-differentiation formula for \(u^v\), by contrast, uses \(\ln u\) and requires a positive base \(u\).
A Common Error: Keeping Only the Exponent's Contribution
A frequent mistake for \(x^x\) is to differentiate only the base as if the exponent were constant, producing \(x\cdot x^{x-1}=x^x\). That calculation omits the change in the exponent. At \(x=2\), it would give \(4\), whereas the correct derivative is \(4(\ln2+1)\).
The correction is to take logarithms and differentiate the entire product \(x\ln x\):
Another related caution is to distinguish the relative rate from the derivative. The relative rate is \(y'/y\); to obtain \(y'\), multiply that quotient by \(y\). This distinction is what makes the final step of each example necessary.
A Reliable Calculation
For a variable-power function \(y=u(x)^{v(x)}\), the calculation can be organized as follows:
For real-valued logarithmic differentiation, verify that \(u(x)>0\) on the interval of interest.
Use \(y'/y=v'\ln u+v(u'/u)\), keeping both the exponent and base contributions.
Substitute the specified input into the relative-rate expression and calculate \(y\) there.
Multiply the relative rate by \(y\), then verify the result by differentiating the exponential form \(y=\exp(v\ln u)\).
Relative rates are useful whenever change is best understood in proportion to the current value. For variable powers, they expose the two distinct sources of change—the base and the exponent—without obscuring either one. The quotient is also meaningful for any differentiable nonzero function, but the real-logarithm method used here requires a positive function value and, for \(u^v\), a positive base.
Check Your Understanding
Use the relative-rate formula and state any domain condition required.
- For \(y=x^x\), what is \(y'/y\) when \(x>0\)?
- For \(y=(x+2)^{3x}\), find \(y'/y\) on the interval where the base is positive.
- If \(f(x)>0\) and \(f'(x)/f(x)<0\), what does this imply about \(f'(x)\) at that point?
- For \(f(x)=-e^x\), calculate \(f'/f\) and compare its sign with the sign of \(f'\).
- For \(y=(x^2+1)^{2x}\), write the relative-rate expression without expanding or simplifying it.
