Question

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

Answer 1

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,

& \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{aligned}$$ Grouping the terms containing $$\dfrac{dy}{dx}$$ together, we get, $$\begin{aligned} & \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{aligned}$$ Cancelling the common factor $$\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)$$.