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If $y=\left(1+x^{\frac{1}{4}}\right)\left(1+x^{\frac{1}{2}}\right)\left(1-x^{\frac{1}{4}}\right)$, then what is $\frac{d y}{d x}$ equal to?
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Verified Answer
The correct answer is:
$-1$
Let $y=\left(1+x^{1 / 4}\right)\left(1+x^{1 / 2}\right)\left(1-x^{1 / 4}\right)$
$=\left(1+x^{1 / 4}\right)\left(1-x^{1 / 4}\right)\left(1+x^{1 / 2}\right)$
$=\left(1-x^{1 / 2}\right)\left(1+x^{1 / 2}\right)$
$\left(\because(a+b)(a-b)=a^{2}-b^{2}\right)$
$\begin{array}{l}
=(1-x)\left(\because(a+b)(a-b)=a^{2}-b^{2}\right) \\
\Rightarrow y=1-x
\end{array}$
Differentiate both side w.r.t ' $x$, $\frac{d y}{d x}=-1$
$=\left(1+x^{1 / 4}\right)\left(1-x^{1 / 4}\right)\left(1+x^{1 / 2}\right)$
$=\left(1-x^{1 / 2}\right)\left(1+x^{1 / 2}\right)$
$\left(\because(a+b)(a-b)=a^{2}-b^{2}\right)$
$\begin{array}{l}
=(1-x)\left(\because(a+b)(a-b)=a^{2}-b^{2}\right) \\
\Rightarrow y=1-x
\end{array}$
Differentiate both side w.r.t ' $x$, $\frac{d y}{d x}=-1$
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