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

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

Explanation

Solution

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 xx.

Formula used: log⁡x=1x\log x = \dfrac{1}{x}
d(xy)=ydx−xdyy2d\left( {\dfrac{x}{y}} \right) = \dfrac{{ydx - xdy}}{{{y^2}}}

Complete step-by-step solution:
We are given an equation –
⇒log⁡(x−y)=xy\Rightarrow \log \left( {x - y} \right) = \dfrac{x}{y} …. (given)
Differentiating both the sides with respect to xx,by using the formula
⇒1−dydxx−y=y−xdydxy2\Rightarrow \dfrac{{1 - \dfrac{{dy}}{{dx}}}}{{x - y}} = \dfrac{{y - x\dfrac{{dy}}{{dx}}}}{{{y^2}}}
Now, we will simplify to find dydx\dfrac{{dy}}{{dx}}.
⇒(1−dydx)y2=(y−xdydx)(x−y)\Rightarrow \left( {1 - \dfrac{{dy}}{{dx}}} \right){y^2} = \left( {y - x\dfrac{{dy}}{{dx}}} \right)\left( {x - y} \right)
Opening the brackets,
⇒y2−y2dydx=xy−y2−x2dydx+xy\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 dydx\dfrac{{dy}}{{dx}} on one side and the other terms on the other side.
⇒y2+y2−2xy=y2dydx−x2dydx\Rightarrow {y^2} + {y^2} - 2xy = {y^2}\dfrac{{dy}}{{dx}} - {x^2}\dfrac{{dy}}{{dx}}
Taking dydx\dfrac{{dy}}{{dx}} common,
⇒2y2−2xy=(y2−x2)dydx\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 dydx\dfrac{{dy}}{{dx}},
⇒2y2−xyy2−x2=dydx\Rightarrow \dfrac{{2{y^2} - xy}}{{{y^2} - {x^2}}} = \dfrac{{dy}}{{dx}}
Now, I can notice that on taking 2y2y common in the numerator, I will be able to simplify my answer further.
⇒2y(y−x)(y+x)(y−x)=dydx\Rightarrow \dfrac{{2y\left( {y - x} \right)}}{{\left( {y + x} \right)\left( {y - x} \right)}} = \dfrac{{dy}}{{dx}} …. (using the identity a2−b2=(a+b)(a−b){a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right))
On simplifying it further, we get,
⇒2yx+y=dydx\Rightarrow \dfrac{{2y}}{{x + y}} = \dfrac{{dy}}{{dx}}

The answer for the given question is 2yx+y=dydx\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 y2{y^2} will be 2ydydx2y\dfrac{{dy}}{{dx}}. If we had to differentiate with respect to y, then in that case, differentiation of y2{y^2} would have been only 2y2y. In this method, we use chain rules to differentiate.