Question

If \[x{{e}^{xy}}-y=\sin x\] then \[\dfrac{dy}{dx}\] at \[x=0\] is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer 1

Hint: We will first differentiate both sides of \[x{{e}^{xy}}-y=\sin x\] implicitly on left hand side and on the right hand side and for this we will use product rule \[\dfrac{d(uv)}{dx}=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\] to differentiate. Then finally we will substitute x equal to 0 in this and get our answer.

Complete step-by-step answer:
The given expression in the question is \[x{{e}^{xy}}-y=\sin x.........(1)\].

So now differentiating with respect to x on both sides of equation (1) and applying product rule for first term of the left hand side of the equation (1), we get,

\[x{{e}^{xy}}\left( 1+\dfrac{dy}{dx} \right)+{{e}^{xy}}-\dfrac{dy}{dx}=\cos x.........(2)\]

Now taking common terms out and rearranging in equation (2) we get,

\[\dfrac{dy}{dx}\left( x{{e}^{xy}}-1 \right)+x{{e}^{xy}}+{{e}^{xy}}=\cos x.........(3)\]

Now again rearranging and isolating \[\dfrac{dy}{dx}\] in equation (3) we get,

\[\dfrac{dy}{dx}=\dfrac{\cos x-x{{e}^{xy}}-{{e}^{xy}}}{x{{e}^{xy}}-1}.........(4)\]

Now the question has asked us to find \[\dfrac{dy}{dx}\] at x equal to 0. So now substituting it in equation (4) we get,

\[\dfrac{dy}{dx}=\dfrac{\cos 0-0{{e}^{0y}}-{{e}^{0y}}}{0{{e}^{0y}}-1}.........(5)\]

Now we know that cos 0 is 1 and hence simplifying all the terms in equation (5) we get,
\[\dfrac{dy}{dx}=\dfrac{1-1}{-1}=0\]

Hence the correct answer is option (a).

Note: Remembering the basic differentiation formulas of some basic functions and the product rule of differentiation is the key here. Also we need to substitute for x in equation (4) to get the final answer. We can make a mistake in equation (5) in a hurry by writing \[{{e}^{0y}}\] as 0 instead of 1 and we may also get confused in doing the implicit differentiation of the left hand side of equation (1).