Question

Differentiate the given function with respect to x; \[{{\log }_{7}}\left( 2x-3 \right)\].

Answer 1

Hint: To solve this question we should know the derivative of \[\log x\].
Derivative of \[\log x\] with respect to x is given by, \[\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x}\]. Also we need chain rule of differentiation. The chain rule of differentiation states that derivatives of \[f\left( g\left( x \right) \right)\] is \[{{f}^{'}}\left( g\left( x \right) \right){{g}^{'}}\left( x \right)\].

Complete step-by-step solution:
Given the function is \[{{\log }_{7}}\left( 2x-3 \right)\]. Because the base of \[\log \] is not ‘e’. So, we first convert it to base e using formula base.
\[{{\log }_{a}}b=\dfrac{{{\log }_{e}}b}{{{\log }_{e}}a}\]
So we can convert \[{{\log }_{7}}\left( 2x-3 \right)\] using above formula as,
\[{{\log }_{7}}\left( 2x-3 \right)=\dfrac{{{\log }_{e}}\left( 2x-3 \right)}{{{\log }_{e}}7}\]
Now because \[\dfrac{1}{{{\log }_{e}}7}\] is a constant value independent of x while deviating this value \[\dfrac{1}{{{\log }_{e}}7}\] comes out common.
Differentiating \[\dfrac{{{\log }_{e}}\left( 2x-3 \right)}{{{\log }_{e}}7}\] with respect to x we get,
\[\dfrac{d}{dx}\left( \dfrac{{{\log }_{e}}\left( 2x-3 \right)}{{{\log }_{e}}7} \right)=\dfrac{1}{{{\log }_{e}}7}\dfrac{d}{dx}\left( {{\log }_{e}}\left( 2x-3 \right) \right)\]
Now we will use chain rule of differentiation to deviate thus,
\[\begin{align}
  & \dfrac{1}{{{\log }_{e}}7}\dfrac{d}{dx}\left( {{\log }_{e}}\left( 2x-3 \right) \right)=\dfrac{1}{{{\log }_{e}}7}\left[ \dfrac{1}{\left( 2x-3 \right)}\dfrac{d}{dx}\left( 2x-3 \right) \right] \\
 & \Rightarrow \dfrac{1}{{{\log }_{e}}7}\dfrac{d}{dx}\left( {{\log }_{e}}\left( 2x-3 \right) \right)=\dfrac{1}{{{\log }_{e}}7\left( 2x-3 \right)}\left( 2-0 \right) \\
\end{align}\]
Hence the derivative of \[{{\log }_{7}}\left( 2x-3 \right)\] is \[\dfrac{2}{\left( {{\log }_{e}}7 \right)\left( 2x-3 \right)}\].
Hence the derivative of \[{{\log }_{7}}\left( 2x-3 \right)\] is \[\dfrac{2}{\left( {{\log }_{e}}7 \right)\left( 2x-3 \right)}\].

Note: The possibility of mistake in this question can be at the point where student have to apply chain rule of differentiation this is important because our function \[{{\log }_{7}}\left( 2x-3 \right)\] do not have \[\log x\] rather have “\[2x-3\]” is place of x, and \[\dfrac{d}{dx}\left( x \right)=1\]. But \[\dfrac{d}{dx}\left( 2x-3 \right)=2\], which differs the answer.