Question

How do you implicitly differentiate $\log \left( {x - y} \right) = \dfrac{x}{y}$ ?

Answer 1

Hint: In this question, we are given an equation in terms of logarithm and we have been asked to differentiate it. The given equation is an implicit equation. At first, read about implicit equations. Then, differentiate the given equation. Though the equation is in terms of log, do not use any logarithmic properties. Simply, differentiate both the sides with respect to $x$.

Formula used: $\log x = \dfrac{1}{x}$
$d\left( {\dfrac{x}{y}} \right) = \dfrac{{ydx - xdy}}{{{y^2}}}$

Complete step-by-step solution:
We are given an equation –
$ \Rightarrow \log \left( {x - y} \right) = \dfrac{x}{y}$ …. (given)
Differentiating both the sides with respect to $x$,by using the formula
$ \Rightarrow \dfrac{{1 - \dfrac{{dy}}{{dx}}}}{{x - y}} = \dfrac{{y - x\dfrac{{dy}}{{dx}}}}{{{y^2}}}$
Now, we will simplify to find $\dfrac{{dy}}{{dx}}$.
$ \Rightarrow \left( {1 - \dfrac{{dy}}{{dx}}} \right){y^2} = \left( {y - x\dfrac{{dy}}{{dx}}} \right)\left( {x - y} \right)$
Opening the brackets,
$ \Rightarrow {y^2} - {y^2}\dfrac{{dy}}{{dx}} = xy - {y^2} - {x^2}\dfrac{{dy}}{{dx}} + xy$
Now, we will bring all the terms containing $\dfrac{{dy}}{{dx}}$ on one side and the other terms on the other side.
$ \Rightarrow {y^2} + {y^2} - 2xy = {y^2}\dfrac{{dy}}{{dx}} - {x^2}\dfrac{{dy}}{{dx}}$
Taking $\dfrac{{dy}}{{dx}}$ common,
$ \Rightarrow 2{y^2} - 2xy = \left( {{y^2} - {x^2}} \right)\dfrac{{dy}}{{dx}}$
Last step is to shift the terms to find the value of $\dfrac{{dy}}{{dx}}$,
$ \Rightarrow \dfrac{{2{y^2} - xy}}{{{y^2} - {x^2}}} = \dfrac{{dy}}{{dx}}$
Now, I can notice that on taking $2y$ common in the numerator, I will be able to simplify my answer further.
$ \Rightarrow \dfrac{{2y\left( {y - x} \right)}}{{\left( {y + x} \right)\left( {y - x} \right)}} = \dfrac{{dy}}{{dx}}$ …. (using the identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$)
On simplifying it further, we get,
$ \Rightarrow \dfrac{{2y}}{{x + y}} = \dfrac{{dy}}{{dx}}$

The answer for the given question is $\dfrac{{2y}}{{x + y}} = \dfrac{{dy}}{{dx}}$

Note: What is implicit differentiation and how to do it?
Under implicit differentiation, the equation given is a mix of two variables (x and y, in this case). One variable is a function if the other variable. For example: If we are differentiating with respect to x, y is our implicit function. In such a case, differentiation of ${y^2}$ will be $2y\dfrac{{dy}}{{dx}}$. If we had to differentiate with respect to y, then in that case, differentiation of ${y^2}$ would have been only $2y$. In this method, we use chain rules to differentiate.