A derivative of logarithmic functions

You already know how to take the derivative of polynomials, trigonometric, and composite functions. In this topic, you are going to learn about logarithmic functions and its derivatives.

Basic formulas

We recall that the natural logarithm of $x$ is a logarithm with the base $e$:

$$\ln x = \log_e x$$A derivative of natural logarithm is$$(\ln x)' = \dfrac{1}{x}$$We know that $$\log_a x = \dfrac{\ln x}{\ln a}$$So, considering that $\ln a$ is a constant, we obtain$$(\log_a x)' = \left( \dfrac{\ln x}{\ln a}\right)' = \dfrac{1}{\ln a} (\ln x)' = \dfrac{1}{\ln a} \cdot \dfrac{1}{x} = \dfrac{1}{x \ln a}$$

Let's practice!

So, let's see how we can combine these formulas with the other rules we have learned. Say we want to find a derivative of $\ln(\sin 5x)$. This is a composite function, hence we need the chain rule. Firstly, we take the derivative of the natural logarithm and get $\dfrac{1}{\sin 5x}$. Secondly, we multiply it by the derivative of what was inside and get:

$$(\ln(\sin 5x))' = \dfrac{1}{\sin 5x} \cdot (\sin 5x)' = \\ \ \\ \dfrac{1}{\sin 5x} \cdot \cos 5x \cdot 5 = \dfrac{5\cos 5x}{\sin 5x} = 5 \cot 5x$$Thus, when we take the derivative of the natural logarithm of any function, like$$f(x) = \ln (g(x))$$we get

$$f'(x) =\dfrac{g'(x)}{g(x)}$$Now let's find the derivative of $\log_{20} x^3$:

$$(\log_{20} x^3)' = \dfrac{1}{x^3 \ln 20} \cdot (x^3)' = \dfrac{3x^2}{x^3 \ln 20} = \dfrac{3}{x\ln 20}$$

An important example

Let's consider a function $f(x)= x^x$. How can we find the derivative of this function? We need to use the technique we have just considered – the logarithmic derivative. We hang logarithms on both sides:

$$\ln f(x) = \ln x^x$$According to the properties of the logarithm, on the right side the degree is taken out from under the logarithm:

$$\ln f(x) = x\ln x$$As a result, on the right-hand side we have a product of two functions, which will be differentiated according to the product rule:

$(h \cdot g)' = h'\cdot g+h \cdot g'$. Let's differentiate both sides of our equality:

$$(\ln f(x))' = (x\ln x)' \ \Rightarrow \\ \ \\ (\ln f(x))' = x' \cdot \ln x + x \cdot (\ln x)' \\ \ \\ \dfrac{1}{f(x)} \cdot f'(x) = x' \cdot \ln x + x \cdot (\ln x)' = \ln x + \dfrac{x}{x} = \ln x + 1$$So, we get

$$f'(x) = f(x) \cdot (\ln x + 1) = x^x \cdot (\ln x + 1)$$

Conclusion

In this topic, we have learned about logarithmic derivatives and done some practice. Also, we have considered an important example of a situation when the logarithmic derivative allows calculating the derivatives of exponential functions.