Question
Differentiate the following functions with respect to x:
$\log_7(2\text{x}-3)$
$\log_7(2\text{x}-3)$
✓
Answer
Let, $\text{y}=\log_7(2\text{x}-3)$
$\Rightarrow\ \text{y}=\frac{\log(2\text{x}-3)}{\log_7}\ \Big[\text{Since}, \log^\text{b}_\text{a}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\log7}\frac{\text{d}}{\text{dx}}\big(\log(2\text{x}-3)\big)$
$=\frac{1}{\log7}\times\frac{1}{(2\text{x}-3)}\frac{\text{d}}{\text{dx}}(2\text{x}-3)$
[Using chain rule]
$=\frac{2}{(2\text{x}-3)\log7}$
Hence, $\frac{\text{d}}{\text{dx}}(\log_7(2\text{x}-3))=\frac{2}{(2\text{x}-3)\log7}$
$\Rightarrow\ \text{y}=\frac{\log(2\text{x}-3)}{\log_7}\ \Big[\text{Since}, \log^\text{b}_\text{a}=\frac{\log\text{b}}{\log\text{a}}\Big]$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\log7}\frac{\text{d}}{\text{dx}}\big(\log(2\text{x}-3)\big)$
$=\frac{1}{\log7}\times\frac{1}{(2\text{x}-3)}\frac{\text{d}}{\text{dx}}(2\text{x}-3)$
[Using chain rule]
$=\frac{2}{(2\text{x}-3)\log7}$
Hence, $\frac{\text{d}}{\text{dx}}(\log_7(2\text{x}-3))=\frac{2}{(2\text{x}-3)\log7}$
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