Introduction
A derivative condition can reveal an unknown parameter in a function. For a variable-power function, that parameter may appear in the exponent, so differentiating first can seem to produce a complicated expression. Logarithmic differentiation provides a systematic way to organize the calculation.
The essential idea is to take the natural logarithm of the function before differentiating. This turns a variable exponent into a factor, and the derivative can then be evaluated at the specified point. The resulting equation may determine the unknown constant directly, provided the derivative condition actually depends on it.
The Derivative Formula for a Variable Power
Consider a function of the form \(y=g(x)^{h_c(x)}\), where \(c\) is an unknown constant. For real-valued logarithmic differentiation, assume \(g(x)>0\) on an interval around the point of interest, and assume \(g\) and \(h_c\) are differentiable there. Since \(y>0\), we may take its natural logarithm:
Differentiate both sides with respect to \(x\). The chain rule gives \(\frac{y'}{y}\) on the left, while the product rule applies on the right. Thus
Multiplying by \(y=g(x)^{h_c(x)}\) gives the derivative formula:
This formula separates two contributions. The term \(h_c'(x)\ln g(x)\) comes from the changing exponent. The term \(h_c(x)g'(x)/g(x)\) comes from the changing base. Both terms are needed unless the relevant derivative is zero.
A Useful Situation for Recovering the Constant
A particularly convenient exponent has the form \(h_c(x)=q(x)+c\phi(x)\), where \(q\) and \(\phi\) are known functions. Suppose the derivative is specified at \(x=x_0\), and \(\phi(x_0)=0\). Then the function value at \(x_0\) does not depend on \(c\). Differentiating the exponent gives \(h_c'(x_0)=q'(x_0)+c\phi'(x_0)\), so the derivative condition is an equation that is linear in \(c\).
More precisely, assuming \(g(x_0)>0\), the formula becomes
The coefficient of \(c\) is \(g(x_0)^{q(x_0)}\phi'(x_0)\ln g(x_0)\). If that coefficient is nonzero, a specified value of \(y'(x_0)\) determines exactly one \(c\). If it is zero, this condition does not determine \(c\): it either fails to hold for any \(c\), or holds for every \(c\) in the family. This observation is useful before doing lengthy algebra.
Worked Examples
Worked Example: Recovering the Constant from a Derivative at \(x=2\)
Let \(f(x)=x^{c(x-2)+2}\) for \(x>0\). Find \(c\) if \(f'(2)=4(4+8\ln 2)\).
Set \(h_c(x)=c(x-2)+2\). Then \(h_c'(x)=c\). Logarithmic differentiation gives
At \(x=2\), the exponent is \(h_c(2)=2\), so \(f(2)=2^2=4\). The logarithmic derivative there is
Therefore the actual derivative is \(f'(2)=4(c\ln 2+1)\). Apply the given condition and divide both sides by \(4\):
Subtract \(1\) from both sides, then divide by \(\ln 2\):
Check the result by substitution into the derivative condition. For this value, \(c\ln 2+1=8\ln 2+3+1=4+8\ln 2\), so \(f'(2)=4(4+8\ln 2)\), as required.
Worked Example: A Changing Base with a Shifted Exponent
Let \(f(x)=(x^2+1)^{c(x-1)+1}\). Find \(c\) if \(f'(1)=6\).
The base \(x^2+1\) is positive for every real \(x\), so logarithmic differentiation applies. Write \(h_c(x)=c(x-1)+1\), for which \(h_c'(x)=c\). Taking logarithms and differentiating gives
At \(x=1\), the exponent is \(1\) and the base is \(2\), so \(f(1)=2\). The two terms in the logarithmic derivative evaluate to \(c\ln 2\) and \(1\cdot\frac{2}{2}=1\). Hence
Set this equal to \(6\):
For the check, substitute \(c=2/\ln 2\) into the evaluated derivative: \(f'(1)=2((2/\ln 2)\ln 2+1)=2(2+1)=6\). This matches the specified value.
Worked Example: Accounting for Both the Exponent and the Base
Let \(f(x)=(x+2)^{c(x-1)+x^2}\). Find \(c\) if
Here the exponent is \(h_c(x)=c(x-1)+x^2\), so \(h_c'(x)=c+2x\). The base is \(g(x)=x+2\), with \(g'(x)=1\). The logarithmic derivative formula gives
At \(x=1\), the exponent is \(1\), and the base is \(3\), so \(f(1)=3\). The logarithmic derivative is
Consequently, \(f'(1)=3\bigl((c+2)\ln 3+\frac{1}{3}\bigr)\). Equating this to the given derivative and dividing by \(3\) gives
Subtract \(\frac{1}{3}\) from both sides, then divide by \(\ln 3\), which is nonzero:
Check by substitution: with \(c=3\), the logarithmic derivative at \(1\) is \(5\ln 3+\frac{1}{3}\), and multiplying by \(f(1)=3\) gives exactly the specified \(f'(1)\).
Why the Method Works
The key step is that taking a logarithm converts the variable power into a product: \(\ln(g^{h_c})=h_c\ln g\). The product rule then exposes how changes in the exponent and base each affect the relative derivative \(f'/f\). Multiplication by \(f\) recovers the ordinary derivative. This also explains why a derivative condition must be applied to \(f'\), not merely to \(f'/f\).
The derivation requires a positive base near the point where the derivative is evaluated. This ensures that the real-valued logarithm \(\ln g(x)\) is defined. It also requires differentiability of the base and exponent. If the base is not positive, this real logarithmic method may not apply, and the function itself may not be real-valued for arbitrary real exponents.
Common Mistakes and How to Avoid Them
A frequent error is to calculate \(f'/f\) and treat it as \(f'\). In the first example, the logarithmic derivative at \(x=2\) is \(c\ln 2+1\), but \(f(2)=4\). Therefore \(f'(2)=4(c\ln 2+1)\), not just \(c\ln 2+1\). Omitting the factor \(4\) changes the equation being solved.
Algebra after differentiation also deserves a deliberate check. From the first example, the equation is \(c\ln 2+1=4+8\ln 2\). Subtracting \(1\) gives \(c\ln 2=3+8\ln 2\), so division by \(\ln 2\) yields \(c=8+3/\ln 2\). The expression \(3+7/\ln 2\) does not follow from that equation. Substituting the proposed value back into the equation is a simple way to catch such an error.
Finally, a derivative condition may fail to identify the parameter if the parameter’s coefficient disappears at the chosen point. Before dividing by a coefficient, check that it is nonzero. A zero coefficient is not permission to divide; it means the condition must be examined separately for consistency.
Summary
To recover an unknown exponent constant from a derivative condition, take the natural logarithm of the variable-power function, differentiate using the product and chain rules, and then evaluate the complete derivative at the specified point. Keep the function-value factor when converting from \(f'/f\) to \(f'\), solve the resulting equation carefully, and substitute the answer back into the condition.
Check Your Understanding
Use logarithmic differentiation where needed. In each problem, show how the derivative condition leads to the value of the constant and check your result.
- For \(f(x)=x^{c(x-1)+1}\), find \(c\) if \(f'(1)=3\).
- For \(f(x)=(x+1)^{c(x-2)+2}\), write a formula for \(f'(2)\) in terms of \(c\).
- For \(f(x)=(x^2+2)^{c(x-1)+1}\), find \(c\) if \(f'(1)=3(2\ln 3+\frac{2}{3})\).
- Why must the function-value factor be included when using a calculated value of \(f'/f\) to find \(f'\)?
- In the form \(h_c(x)=q(x)+c\phi(x)\), what does it imply for recovering \(c\) if the coefficient of \(c\) in the evaluated derivative condition is zero?
