Question

How do you find the derivative of \[\dfrac{\ln x}{\ln y}=\ln \left( x-y \right)\]?

Answer 1

Hint: Multiply both the sides of the given equation with \[\ln y\] to get rid of the fraction. Now, consider the R.H.S. of the simplified equation as the product of two logarithmic functions and assume them as u and v respectively. Now, apply the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\] to simplify the R.H.S. and get the answer. Use the formula: - \[\dfrac{d\ln x}{dx}=\dfrac{1}{x}\].

Complete step by step answer:
Here, we have been provided with the function \[\dfrac{\ln x}{\ln y}=\ln \left( x-y \right)\] and we are asked to find its derivative. That means we have to find the value of \[\dfrac{dy}{dx}\].
\[\because \dfrac{\ln x}{\ln y}=\ln \left( x-y \right)\]
Multiplying both the sides with \[\ln y\], we get,
\[\Rightarrow \ln x=\ln y\ln \left( x-y \right)\]
Here, we can consider the R.H.S. of the above expression as the product of two logarithmic functions. Let us assume \[\ln y=u\] and \[\ln \left( x-y \right)=v\], so we have,
\[\Rightarrow \ln x=u\times v\]
Differentiating both the sides with respect to x, we get,
\[\Rightarrow \dfrac{d\left( \ln x \right)}{dx}=\dfrac{d\left( u\times v \right)}{dx}\]
Now, using the product rule of differentiation given as: - \[\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\], we get,
\[\Rightarrow \dfrac{d\left( \ln x \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}\]
Substituting the assumed values of u and v, we have,
\[\Rightarrow \dfrac{d\left( \ln x \right)}{dx}=\ln y\left[ \dfrac{d\ln \left( x-y \right)}{dx} \right]+\ln \left( x-y \right)\left[ \dfrac{d\ln y}{dx} \right]\]
The above expression can be written using the chain rule of differentiation as: -
\[\Rightarrow \dfrac{d\ln x}{dx}=\ln y\left[ \dfrac{d\ln \left( x-y \right)}{d\left( x-y \right)}\times \dfrac{d\left( x-y \right)}{dx} \right]+\ln \left( x-y \right)\left[ \dfrac{d\ln y}{dy}\times \dfrac{dy}{dx} \right]\]
Using the formula: - \[\dfrac{d\ln x}{dx}=\dfrac{1}{x}\], we have,
\[\begin{align}
  & \Rightarrow \dfrac{1}{x}=\ln y\left[ \dfrac{1}{\left( x-y \right)}\times \left( \dfrac{dx}{dx}-\dfrac{dy}{dx} \right) \right]+\ln \left( x-y \right)\left[ \dfrac{1}{y}\dfrac{dy}{dx} \right] \\
 & \Rightarrow \dfrac{1}{x}=\dfrac{\ln y}{\left( x-y \right)}\left( 1-\dfrac{dy}{dx} \right)+\dfrac{\ln \left( x-y \right)}{y}\dfrac{dy}{dx} \\
\end{align}\]
Grouping the terms containing \[\dfrac{dy}{dx}\] together, we get,
\[\begin{align}
  & \Rightarrow \dfrac{1}{x}=\dfrac{\ln y}{\left( x-y \right)}+\dfrac{dy}{dx}\left( \dfrac{\ln \left( x-y \right)}{y}-\dfrac{\ln y}{\left( x-y \right)} \right) \\
 & \Rightarrow \dfrac{dy}{dx}\left( \dfrac{\left( x-y \right)\ln \left( x-y \right)-y\ln y}{y\left( x-y \right)} \right)=\dfrac{1}{x}-\dfrac{\ln y}{\left( x-y \right)} \\
 & \Rightarrow \dfrac{dy}{dx}\left( \dfrac{\left( x-y \right)\ln \left( x-y \right)-y\ln y}{y\left( x-y \right)} \right)=\left( \dfrac{\left( x-y \right)-x\ln y}{x\left( x-y \right)} \right) \\
\end{align}\]
Cancelling the common factor \[\left( x-y \right)\] from both the sides and simplifying, we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}\left( \dfrac{\left( x-y \right)-x\ln y}{\left( x-y \right)\ln \left( x-y \right)-y\ln y} \right)\]
Hence, the above obtained relation is our answer.

Note: One may note that the given function is an implicit function, that means the variables cannot be separated. We cannot write the variable y in terms of x alone. This is the reason we are getting terms of y in the expression of \[\dfrac{dy}{dx}\]. Note that here we have applied the product rule of differentiation to solve the question. You can apply the \[\dfrac{u}{v}\] rule to solve the question. In that case we would not be required to multiply both the sides with \[\ln y\] at the initial step of the solution. Remember the formula: - \[\dfrac{d\left( \dfrac{u}{v} \right)}{dx}=\left( \dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}} \right)\].