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DA5. Logarithmic Differentiation

Statement

Why might one prefer to use the logarithmic differentiation for finding derivatives of functions that have another function as an exponent?

Solution

For functions of shape \(f(x) = x^n\) we can use the power rule to find the derivative. However, if we have a function of shape \(f(x) = g(x)^{h(x)}\) we cannot use the power rule to find the derivative as the exponent is not constant. Instead, we can use logarithmic differentiation to find the derivative.

The logarithmic differentiation is defined as follows:

\[ \begin{aligned} f(x) &= g(x)^{h(x)} \\ \ln(f(x)) &= \ln(g(x)^{h(x)}) \text{ ---(1)}\\ \ln(f(x)) &= h(x) \ln(g(x)) \text{ ---(2)} \\ \frac{f'(x)}{f(x)} &= h'(x) \ln(g(x)) + \frac{h(x) g'(x)}{g(x)} \text{ ---(3)}\\ f'(x) &= f(x) \left( h'(x) \ln(g(x)) + \frac{h(x) g'(x)}{g(x)} \right) \text{ ---(4)} \\ \end{aligned} \]

Considerations:

  1. By taking the natural logarithm of both sides of the equation, we can get rid of the exponent and then we can utilize other rules to find the derivative.
  2. According to the logarithm rules, we can move the exponent down and multiply it with the logarithm of the base. such that \(\ln(a^b) = b \ln(a)\).
  3. The product rule is used to find the derivative of the right side of the equation.
  4. We can multiply both sides of the equation by \(f(x)\) to get the derivative of \(f(x)\).

As an example, let’s find the derivative of \(f(x) = x^x\) using logarithmic differentiation (The Organic Chemistry Tutor, 2016):

\[ \begin{aligned} f(x) &= x^x \\ \ln(f(x)) &= \ln(x^x) \\ \ln(f(x)) &= x \ln(x) \\ \frac{f'(x)}{f(x)} &= \ln(x) + 1 \\ f'(x) &= f(x) \left( \ln(x) + 1 \right) \\ f'(x) &= x^x \left( \ln(x) + 1 \right) \\ \end{aligned} \]

Another example is to find the derivative of \(f(x) = (\ln (x)) ^ x\), using logarithmic differentiation (The Organic Chemistry Tutor, 2016):

\[ \begin{aligned} f(x) &= (\ln (x)) ^ x \\ \ln(f(x)) &= \ln((\ln (x)) ^ x) \\ \ln(f(x)) &= x \ln(\ln (x)) \\ \frac{f'(x)}{f(x)} &= \ln(\ln (x)) + \frac{x}{\ln (x)} \cdot \frac{1}{x} \\ f'(x) &= f(x) \left( \ln(\ln (x)) + \frac{1}{\ln (x)} \right) \\ f'(x) &= (\ln (x)) ^ x \left( \ln(\ln (x)) + \frac{1}{\ln (x)} \right) \\ \end{aligned} \]

References

  • The Organic Chemistry Tutor. (2016, October 14). Logarithmic differentiation Video. YouTube.