If y=y(x) and it follows the relation 4x{e^{x | vedclass

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(B) Given the relation: $4x{e^{xy}} = y + 5{\sin ^2}x$. First,find the value of $y$ at $x=0$. Substituting $x=0$ into the equation: $4(0){e^{0}} = y +

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$4$

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(B) Given the relation: $4x{e^{xy}} = y + 5{\sin ^2}x$. First,find the value of $y$ at $x=0$. Substituting $x=0$ into the equation: $4(0){e^{0}} = y + 5{\sin ^2}(0) \implies 0 = y + 0 \implies y(0) = 0$. Now,differentiate both sides with respect to $x$ using the product rule and chain rule: $\frac{d}{dx}(4x{e^{xy}}) = \frac{d}{dx}(y + 5{\sin ^2}x)$ $4{e^{xy}} + 4x{e^{xy}}(y + x y') = y' + 10\sin x \cos x$. Substitute $x=0$ and $y=0$ into the differentiated equation: $4{e^{0}} + 4(0){e^{0}}(0 + 0 \cdot y'(0)) = y'(0) + 10\sin(0)\cos(0)$. $4(1) + 0 = y'(0) + 0$. Therefore,$y'(0) = 4$.

If $y=y(x)$ and it follows the relation $4x{e^{xy}} = y + 5{\sin ^2}x$,then $y'(0)$ is equal to

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