Question

If \[x{{e}^{xy}}+y{{e}^{-xy}}={{\sin }^{2}}x\] , then \[\dfrac{dy}{dx}\]at x=0 is

Answer 1

HINT: -
As in the equation, it can be seen that the equation consists of different multivariable or two variable functions (to be precise) and differentiating them is a little different from the normal differentiation.

Complete step-by-step answer:

For differentiating a two variable function, we can use the following method to find the differentiation that is \[\dfrac{dy}{dx}\]at x=0 which is required.

The most important formula that would be used in solving this question and getting to the correct solution is as follows

\[df(x,y)=\dfrac{\partial f(x,y)}{\partial x}dx+\dfrac{\partial f(x,y)}{\partial y}dy\]

As mentioned in the question, we have to find the value of \[\dfrac{dy}{dx}\]at x=0.

Now, using the formula given in the hint on every function present in the equation, we can write the following

\[\begin{align}

  & x{{e}^{xy}}+y{{e}^{-xy}}={{\sin }^{2}}x \\

 & dx{{e}^{xy}}+dy{{e}^{-xy}}=d{{\sin }^{2}}x \\

 & \dfrac{\partial x{{e}^{xy}}}{\partial x}dx+\dfrac{\partial x{{e}^{xy}}}{\partial y}dy+\dfrac{\partial y{{e}^{-xy}}}{\partial x}dx+\dfrac{\partial y{{e}^{-xy}}}{\partial y}dy=\dfrac{\partial {{\sin }^{2}}x}{\partial x}dx+\dfrac{\partial {{\sin }^{2}}x}{\partial y}dy \\
 & \left( xy{{e}^{xy}}+{{e}^{xy}} \right)dx+\left( {{x}^{2}}{{e}^{xy}} \right)dy+\left( -{{y}^{2}}{{e}^{-xy}} \right)dx+\left( -xy{{e}^{-xy}}+{{e}^{-xy}} \right)dy=\left( 2\sin x\cdot \cos x \right)dx+0 \\

\end{align}\]

Now, putting x=0, we will get the following result which is as follows

\[\begin{align}

  & \left( 0y{{e}^{0y}}+{{e}^{0y}} \right)dx+\left( {{0}^{2}}{{e}^{0y}} \right)dy+\left( -{{y}^{2}}{{e}^{-0y}} \right)dx+\left( -0y{{e}^{-0y}}+{{e}^{-0y}} \right)dy=\left( 2\sin 0\cdot \cos 0 \right)dx+0 \\

 & dx+\left( -{{y}^{2}} \right)dx+dy=0 \\

 & dy=\left( {{y}^{2}}-1 \right)dx \\

 & \dfrac{dy}{dx}=\left( {{y}^{2}}-1 \right) \\

\end{align}\]

Hence, this is the value of \[\dfrac{dy}{dx}\]at x=0.


NOTE: -

The students can make an error if they don’t know how to differentiate this particular equation as it is a very tricky one to differentiate and it is not easy to simply differentiate this equation and get the required \[\dfrac{dy}{dx}\]at x=0.

The most important formula that would be used in solving this question is as follows which is mentioned in the hint as well

\[df(x,y)=\dfrac{\partial f(x,y)}{\partial x}dx+\dfrac{\partial f(x,y)}{\partial y}dy\] .