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Algebra of Derivatives

If y = 1 + ...

Question

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

A

1

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B

- 1

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C

x

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D

\sqrt{x}

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Solution

The correct option is A $- 1$
$y = (1 + x^{1 / 4}) (1 + x^{1 / 2}) (1 - x^{1 / 4}) = (1 - x^{1 / 2}) (1 + x^{1 / 2})$
$⟹ y = 1 - x$
$\frac{d y}{d x} = 0 - 1 = - 1$

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Similar questions

y = log ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ {(\frac{1 + x}{1 - x})}^{\frac{1}{4}} ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ - \frac{1}{2} {tan}^{- 1} (x)

\frac{d y}{d x} =

y = log {(\frac{1 + x}{1 - x})}^{\frac{1}{4}} - \frac{1}{2} {tan}^{- 1} (x)

\frac{d y}{d x} =

y = (1 - x) (1 + x^{2}) (1 + x^{4})

\frac{d y}{d x}

y = (1 + x) (1 + x^{2}) (1 + x^{4}) . . . (1 + x^{2^{n}})

\frac{d y}{d x}

x = 0

y = (1 + x) (1 + x^{2}) (1 + x^{4}) \dots (1 + x^{2^{n}})

\frac{d y}{d x}

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