Question

Explain how we can find the derivative of the given equation;
$x{e^y} + y = xy$

Answer 1

Hint : Read the given equation well. Looking at it, we should be able to identify the way in which that equation must be differentiated. It can be done easily using implicit methods of differentiation. When we say implicit differentiation it simply means that we leave the function as it is, rather than separating any variable, so that we can differentiate it directly and then later separate the required derivative.

Complete step-by-step answer :
The way to go about this question is easy. First write down the equation, then differentiate with respect to $x$ , then using the formulas used in differentiation just expand the terms, then finally remember to isolate $\dfrac{{dy}}{{dx}}$ , then we can get the required answer.

Given here the original equation whose derivative we wish to find:
$\Rightarrow x{e^y} + y = xy$
Taking derivative on both sides gives;
$\Rightarrow \dfrac{d}{{dx}}(x{e^y} + y) = \dfrac{d}{{dx}}(xy)$
Expanding using the laws we know;
$\Rightarrow [x.\dfrac{d}{{dx}}({e^y}) + {e^y}.\dfrac{d}{{dx}}(x)] + \dfrac{d}{{dx}}(y) = x.\dfrac{d}{{dx}}(y) + y.\dfrac{d}{{dx}}(x)$
Remember here that chain rule is used so: $\Rightarrow \dfrac{d}{{dx}}({e^y}) = \dfrac{d}{{dx}}({e^y}).\dfrac{{dy}}{{dx}} = {e^y}.\dfrac{{dy}}{{dx}}$
Now solving all the derivatives that were trivial and also substituting what we found in the previous step, we get;
$\Rightarrow [x.{e^y}.\dfrac{{dy}}{{dx}} + {e^y}.1] + \dfrac{{dy}}{{dx}} = x.\dfrac{{dy}}{{dx}} + y.1$
Now in order to perform implicit differentiation, we need to bring all the terms which have $\dfrac{{dy}}{{dx}}$ in it on one side, so here we bring it to the left hand side as follows;
$\Rightarrow x.{e^y}.\dfrac{{dy}}{{dx}} + \dfrac{{dy}}{{dx}} - x.\dfrac{{dy}}{{dx}} = y - {e^y}$
Grouping all terms with $\dfrac{{dy}}{{dx}}$ , gives;
$\Rightarrow [x.{e^y} + 1 - x].\dfrac{{dy}}{{dx}} = y - {e^y}$
Now clearly we can isolate the $\dfrac{{dy}}{{dx}}$ , to find the final answer;
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y - {e^y}}}{{[x.{e^y} + 1 - x]}}$
We can simplify it more by grouping terms with $x$ ;
$\therefore \dfrac{{dy}}{{dx}} = \dfrac{{y - {e^y}}}{{x.[{e^y} - 1] + 1}}$
So, the correct answer is “ $\dfrac{{y - {e^y}}}{{x.[{e^y} - 1] + 1}}$ ”.

Note : There is no particular difference in using implicit or explicit differentiation for differentiating. Since in both the ways the final answer we get is one and the same. In implicit we leave the question as it is and proceed to differentiating but in explicit, we must try to bring the dependent variable to a side separately, and then continue to differentiate. In a linear equation we can easily convert to implicit.