Question

How do you find the derivative of $\log(8x – 1)$ ?

Answer 1

In this question, we need to find the derivative of $\log(8x – 1)$ . We can use the chain rule of differentiation to find the derivative of $\log(8x – 1)$ . A derivative is defined as a rate of change of function with respect to an independent variable given in the function. First, we can consider $u = (8x – 1)$ . Then we need to find the differentiation of $u$ with respect to $x$ and $y$ with respect to $u$ . Then on substituting the values in chain rule and simplifying, we can find the derivative .

Complete step by step solution:
Chain rule :
Chain rule is the derivative of the composite function which is the product of the derivative of the first function and derivative of the second function of the composite function. The use of chain rule is to find the derivative of the composite function.
$\dfrac{dy}{dx} = \dfrac{du}{dx} \times \dfrac{dy}{du}$
Where, $\dfrac{dy}{dx}$ is the derivative of $y$ with respect to $x$
$\dfrac{du}{dx}$ is the derivative of $u$ with respect to $x$
$\dfrac{dy}{du}$ is the derivative of $y$ with respect to $u$
Derivative rules used :
1. $\dfrac{d}{dx}\left( \log x \right) = \dfrac{1}{x}$
2. $\dfrac{d}{dx}\left( kx \right) = k$
3. $\dfrac{d}{dx}\left( k \right) = 0$
Where $k$ is the constant
Complete step by step solution :
Given, $\log(8x – 1)$
Let us consider the given expression as $y$.
$\Rightarrow \ y = \log(8x – 1)$
Here we will use chain rule to find the derivative of $y = \log(8x – 1)$
Chain rule :
$\dfrac{dy}{dx} = \dfrac{du}{dx} \times \dfrac{dy}{du}$
$\Rightarrow \ y = \log(8x – 1)$
Let us consider $u = (8x – 1)$
$\Rightarrow \ y = \log(u)$
On differentiating $u = (8x – 1)$ ,
We get,
$\Rightarrow \dfrac{du}{dx} = 8$
Now we can differentiate $y$ ,
On differentiating both sides with respect to $u$ ,
$\Rightarrow \dfrac{dy}{du} = \dfrac{d}{dx}\left( \log\left( u \right) \right)$
We know that
$\dfrac{d}{dx}\left( \log\ {x} \right) = \dfrac{1}{x}$
Thu we get,
$\Rightarrow \dfrac{dy}{du} = \dfrac{1}{u}$
Here $\dfrac{du}{dx} = 8$ and $\dfrac{dy}{du} = \dfrac{1}{u}$ ,
By substituting the values in the chain rule formula,
$\Rightarrow \dfrac{dy}{dx} = \dfrac{du}{dx} \times \dfrac{dy}{du}$
We get,
$\dfrac{dy}{dx} = 8 \times \dfrac{1}{u}$
By substituting, $u = (8x – 1)$
We get,
$\Rightarrow \dfrac{dy}{dx} = \dfrac{8}{8x – 1}$
Thus we get the derivative of $\log\left( 8x – 1 \right)$ is $\dfrac{8}{8x – 1}$ .
The derivative of $\log\left( 8x – 1 \right)$ is $\dfrac{8}{8x – 1}$ .

Note:
This problem deals with differentiation and logarithms. In the process of answering the questions of this type we should be sure with the concepts that we are going to apply in between the steps. We need to note that while solving the problem here we used the chain rule of differentiation . We should not make any calculation mistakes while solving the question and while doing differentiation, we need to do the differentiation carefully and systematically .