Question

Find dy/dx if x + sin (xy)-y = 0​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Given,

$\[x + \sin \left( {xy} \right) - y = 0\]$

Solution,

$\[x + \sin \left( {xy} \right) - y = 0\]$

$\[ \Rightarrow 1 + \cos \left( {xy} \right)\left( {x\frac{{dy}}{{dx}} + y} \right) - \frac{{dy}}{{dx}} = 0\]$

$\[ \Rightarrow 1 + x\cos \left( {xy} \right)\frac{{dy}}{{dx}} + y\cos \left( {xy} \right) - \frac{{dy}}{{dx}} = 0\]$

$\[ \Rightarrow \frac{{dy}}{{dx}}\left( {x\cos \left( {xy} \right) - 1} \right) + 1 + y\cos \left( {xy} \right) = 0\]$

$\[ \Rightarrow \frac{{dy}}{{dx}} = \frac{{ - \left( {1 + y\cos \left( {xy} \right)} \right)}}{{\left( {x\cos \left( {xy} \right) - 1} \right)}}\]$

$\[ \Rightarrow \frac{{dy}}{{dx}} = \frac{{\left( {1 + y\cos \left( {xy} \right)} \right)}}{{\left( {1 - x\cos \left( {xy} \right)} \right)}}\]$

Hence the value of $\[\frac{{dy}}{{dx}}\]$ is $\[\frac{{\left( {1 + y\cos \left( {xy} \right)} \right)}}{{\left( {1 - x\cos \left( {xy} \right)} \right)}}\]$