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  4. If xexy=y+sin2x, then (dy/dx) at x=0 is:

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Q. If $ x{{e}^{xy}}=y+si{{n}^{2}}x, $ then $ \frac{dy}{dx} $ at $ x=0 $ is:

KEAMKEAM 2006

Solution:

Give that, $ x{{e}^{xy}}=y+si{{n}^{2}}x $ at $ x=0,\text{ }y=0 $ On differentiating w.r.t. $ x, $ we get $ x{{e}^{xy}}\left( x\frac{dy}{dx}+y \right)+{{e}^{xy}}=\frac{dy}{dx}+2\sin x\,\cos x $ $ \Rightarrow $ $ {{x}^{2}}{{e}^{xy}}\frac{dy}{dx}-\frac{dy}{dx}=2\sin x\cos x $ $ -{{e}^{xy}}-x{{e}^{xy}}y $ $ \Rightarrow $ $ \frac{dy}{dx}=\frac{\sin 2x-{{e}^{xy}}(1+xy)}{({{x}^{2}}{{e}^{xy}}-1)} $ $ \therefore $ $ {{\left. \frac{dy}{dx} \right|}_{x=0}}=\frac{0-1(1)}{(-1)}=1 $