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(D) The circuit consists of a $NAND$ gate and an $OR$ gate whose outputs are fed into an $AND$ gate.Let the inputs be $A$ and $B$.The output of the $N

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$X$-$OR$ gate.

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(D) The circuit consists of a $NAND$ gate and an $OR$ gate whose outputs are fed into an $AND$ gate.Let the inputs be $A$ and $B$.The output of the $NAND$ gate is $Y_1 = \overline{A \cdot B}$.The output of the $OR$ gate is $Y_2 = A + B$.These are inputs to the final $AND$ gate,so the final output $Y$ is:$Y = Y_1 \cdot Y_2 = (\overline{A \cdot B}) \cdot (A + B)$Using De Morgan's theorem,$\overline{A \cdot B} = \overline{A} + \overline{B}$.$Y = (\overline{A} + \overline{B}) \cdot (A + B)$$Y = \overline{A} \cdot A + \overline{A} \cdot B + \overline{B} \cdot A + \overline{B} \cdot B$Since $\overline{A} \cdot A = 0$ and $\overline{B} \cdot B = 0$:$Y = 0 + \overline{A} \cdot B + A \cdot \overline{B} + 0$$Y = A \cdot \overline{B} + \overline{A} \cdot B$This is the Boolean expression for an $X$-$OR$ gate.

The logic gate represented by the following logic circuit is

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