Question

Solve the following question in detail:
How do you find \[\dfrac{{dy}}{{dx}}\] by implicit differentiation of \[y = \sin \left( {xy} \right)?\]

Answer 1

Hint: Substitute the given question in the solution we need to find. Then use the rules of the derivative functions to derive the functions. We use the chain rule and the product rule to simplify the equation. Then we bring all the terms of variables to be found on one side and found on the other side. We simplify them until we get the final answer.

Complete step-by-step solution:
Given function,
\[y = \sin \left( {xy} \right)\]
To find \[\dfrac{{dy}}{{dx}}\], we substitute the \[y\] with the given value in the question.
\[ \Rightarrow \dfrac{d}{{dx}}y\]
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin \left( {xy} \right)} \right)\]
Now, using the chain rule of derivatives, we get;
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\sin \left( {xy} \right)} \right) = \left( {\cos \left( {xy} \right)} \right)\dfrac{d}{{dx}}\left( {xy} \right)\]
Using the product rule of derivatives, we get;
\[ \Rightarrow \left\{ {x \times \dfrac{d}{{dx}}y + y \times \dfrac{d}{{dx}}x} \right\}\cos \left( {xy} \right)\]
Multiplying the outside parentheses term with the terms inside, we get;
\[ \Rightarrow x\cos \left( {xy} \right)\dfrac{{dy}}{{dx}} + y\cos \left( {xy} \right)\]
Thus we conclude that\[\dfrac{{dy}}{{dx}} = x\cos \left( {xy} \right)\dfrac{{dy}}{{dx}} + y\cos \left( {xy} \right)\]
Now, taking the like terms to one side of the equation, we get;
\[ \Rightarrow \dfrac{{dy}}{{dx}} - x\cos \left( {xy} \right)\dfrac{{dy}}{{dx}} = y\cos \left( {xy} \right)\]
Taking the common terms out, we get;
\[ \Rightarrow \left( {1 - x\cos \left( {xy} \right)} \right)\dfrac{{dy}}{{dx}} = y\cos \left( {xy} \right)\]
Now, we keep the \[\dfrac{{dy}}{{dx}}\] on one side of the equation; bring the rest of the terms to the other side of the equation. We get;
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{y\cos \left( {xy} \right)}}{{1 - \cos \left( {xy} \right)}}\]

Therefore, we have, \[\dfrac{{dy}}{{dx}} = \dfrac{{y\cos \left( {xy} \right)}}{{1 - x\cos \left( {xy} \right)}}\]

Note: Chain rule is mainly used in calculus, the derivative function. In this rule, the given question has two functions.
While differentiating the functions, we keep one function as a constant and differentiate the other and then we keep the other function as the constant while differentiating the present function. The sum of the two functions is the final derivative answer.
It can also be written in Leibniz’s notation and many other ways. This rule is only used when there are two functions provided for us to differentiate.