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If $ y = x\sin y $ , prove that $ \dfrac{{dy}}{{dx}} = \dfrac{{\sin y}}{{\left( {1 - x\cos y} \right)}} $ .

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Last updated date: 19th Jul 2024
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Answer
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Hint: In this question, we need to prove that $ \dfrac{{dy}}{{dx}} = \dfrac{{\sin y}}{{\left( {1 - x\cos y} \right)}} $ . Here, we will use the given condition $ y = x\sin y $ , and differentiate it with respect to $ x $ . The RHS is in the form of $ u.v $ i.e., the product rule for differentiating equations; therefore we will use the $ u.v $ method here. And further evaluate the terms, we will prove the required condition.

Complete step-by-step answer:
Here, we need to prove that $ \dfrac{{dy}}{{dx}} = \dfrac{{\sin y}}{{\left( {1 - x\cos y} \right)}} $ .
It is given that,
 $ y = x\sin y $
Differentiating both sides with respect to $ x $ , we have,
 $ \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {x\sin y} \right) $
We know that, the product rule for differentiating equation in the form of $ u.v $ is,
 $ \dfrac{d}{{dx}}\left( {u.v} \right) = u.\left( {\dfrac{d}{{dx}}\left( v \right)} \right) + v.\left( {\dfrac{d}{{dx}}\left( u \right)} \right) $
Therefore, $ \dfrac{{dy}}{{dx}} = x.\left( {\dfrac{d}{{dx}}\left( {\sin y} \right)} \right) + \sin y.\left( {\dfrac{d}{{dx}}\left( x \right)} \right) $
Since, we know that $ \dfrac{d}{{dx}}\left( {\sin y} \right) = \cos y $ and $ \dfrac{d}{{dx}}\left( x \right) = 1 $ .
Therefore, by substituting, we have,
 $ \dfrac{{dy}}{{dx}} = x\cos y\left( {\dfrac{{dy}}{{dx}}} \right) + \sin y $
Grouping $ \dfrac{{dy}}{{dx}} $ at one side,
 $ \dfrac{{dy}}{{dx}} - x\cos y\left( {\dfrac{{dy}}{{dx}}} \right) = \sin y $
Taking $ \dfrac{{dy}}{{dx}} $ as common,
 $ \dfrac{{dy}}{{dx}}\left( {1 - x\cos y} \right) = \sin y $
Therefore, $ \dfrac{{dy}}{{dx}} = \dfrac{{\sin y}}{{\left( {1 - x\cos y} \right)}} $
Hence, we have proved $ \dfrac{{dy}}{{dx}} = \dfrac{{\sin y}}{{\left( {1 - x\cos y} \right)}} $ .
So, the correct answer is “ $ \dfrac{{\sin y}}{{\left( {1 - x\cos y} \right)}} $ ”.

Note: In this question, it is important to note here that we can use the product rule in such types of questions. The product rule is nothing but a formula that is used to find the derivatives of products of two or more functions. The rule may be extended to many other solutions, including to products of multiple functions, to a rule for higher-order derivatives of a product, and to other contexts. Here, in case instead of $ y = x\sin y $ , if it is given as $ y = \dfrac{x}{{\sin y}} $ , then can use the quotient rule, it is a method of finding the derivative of a function that is the ratio of two differentiable functions i.e., $ \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v.\left( {\dfrac{d}{{dx}}\left( u \right)} \right) - u.\left( {\dfrac{d}{{dx}}\left( v \right)} \right)}}{{{v^2}}} $ by which we can get the required solution. Be careful while differentiating the terms because it is the area where the mistakes can take place.