Question

What is the derivative of \[y = \log \left( {2x} \right)\]?

Answer 1

Hint: Here we have y as a composition of two functions. We will use the chain rule and other commonly known derivatives of functions to find the derivative of y.

Complete step-by-step solution:
We will first write down the function given in the question, which is,
$y = \log \left( {2x} \right)$ - - - - - - - - - - - - - - - - - - - (1)
This is of the form $y = f(x)$. Now, it is easy to find the derivative of y. We can simply differentiate both sides to get the value of the derivative $\dfrac{{dy}}{{dx}}$.
But before that we have to note a few things about the given function. Observe that $y = \log \left( {2x} \right)$ is a composition of the functions where the outer function is $g(x) = \log x$ and the inner function is $h(x) = 2x$, that is $y = \log \left( {2x} \right) = g\left( {h(x)} \right)$.
Now, if there are two or more functions in composition, to derive them we have to use the chain rule. According to this rule the derivative of composition of two functions $\dfrac{d}{{dx}}g\left( {h(x)} \right) = g'\left( {h(x)} \right)h'(x)$.
Now we will differentiate the given equation (1) on both sides with respect to x to get,
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\log \left( {2x} \right)$
Here we can use the chain rule to get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2x}}.\dfrac{d}{{dx}}\left( {2x} \right)$
[ Using $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$]
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{2x}}.2$ [ Using $\dfrac{d}{{dx}}(ax) = a$]
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{x}$
Hence the derivative of \[y = \log \left( {2x} \right)\] is $\dfrac{1}{x}$.

Note: We can easily get confused when trying to identify the inner and outer functions in a composition of two functions. For example, in ${\cos ^2}(x)$, the outer function is the square function ${x^2}$ and the inner function is $\cos x$.