Question

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

Answer 1

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,

& \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{aligned}$$ Hence the derivative of $${{\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.