If y = (1 + x^{1/4})(1 + x^{1/2})(1 - x^{1/4} | vedclass

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(B) Given the expression: $y = (1 + x^{1/4})(1 + x^{1/2})(1 - x^{1/4})$Rearranging the terms using the commutative property of multiplication:$y = (1

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$-1$

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(B) Given the expression: $y = (1 + x^{1/4})(1 + x^{1/2})(1 - x^{1/4})$Rearranging the terms using the commutative property of multiplication:$y = (1 + x^{1/4})(1 - x^{1/4})(1 + x^{1/2})$Using the algebraic identity $(a + b)(a - b) = a^2 - b^2$,where $a = 1$ and $b = x^{1/4}$:$y = (1^2 - (x^{1/4})^2)(1 + x^{1/2})$$y = (1 - x^{1/2})(1 + x^{1/2})$Again,using the identity $(a - b)(a + b) = a^2 - b^2$,where $a = 1$ and $b = x^{1/2}$:$y = 1^2 - (x^{1/2})^2$$y = 1 - x$Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(1 - x)$$\frac{dy}{dx} = 0 - 1 = -1$

If $y = (1 + x^{1/4})(1 + x^{1/2})(1 - x^{1/4})$,then $\frac{dy}{dx} = $

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