Question

What is the derivative of $\ln \left( 4x \right)$?

Answer 1

For solving this question you should know about how to solve the derivative of a composite function, because $\ln \left( 4x \right)$ is a composite function. And this is a logarithmic function too. So, for solving this we will divide it in two parts and then differentiate and then at the end substitute the values in the chain rule. And thus, we get our answer.

Complete step by step solution:
According to our question we have to find the value of the derivative of $\ln \left( 4x \right)$. So, if we see here, then it is clear that $\ln \left( 4x \right)$ contains two functions, logarithmic function and a normal variable. So, if we consider both of them as two functions, then we will use the chain rule here. And according to the chain rule, we can write $\dfrac{dy}{dx}$ as $\dfrac{dy}{du}.\dfrac{du}{dx}$ and here the $u$ is the second considered function. And so, we will first solve both differentiations and then we solve their value and then put that value in the chain rule again. Then we will find the final value.
So, according to our question$\ln \left( 4x \right)$ if we consider,
$4x=u\ldots \ldots \ldots \left( i \right)$
Then the new function will be,
$y=\ln \left( u \right)$
Now the chain rule is given as,
$\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}$
So, if we calculate $\dfrac{dy}{du}$ , then,
$\begin{aligned} & \dfrac{dy}{du}=\ln \left( u \right) \\\ & \Rightarrow \dfrac{dy}{du}=\dfrac{1}{u} \\\ \end{aligned}$
And,
$\dfrac{du}{dx}=\dfrac{d}{dx}\left( 4x \right)=4$
So, if we now substitute both in the chain rule, then we get,
$\dfrac{dy}{dx}=\dfrac{1}{u}.4$
And $u=4x$ by equation (i), so,
$\dfrac{dy}{dx}=\dfrac{1}{4x}\left( 4 \right)=\dfrac{1}{x}$

So, the derivative of $\ln \left( 4x \right)$ is equal to $\dfrac{1}{x}$.

Note: If you find the derivative of any composite function, then always apply the chain rule there because it makes our calculations easy and accurate. And if we don’t apply this then we have to differentiate the composite function very carefully. And every term should be differentiated.