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Derivative of Standard Functions

If y = 1+x1+x...

Question

If $y$ = $(1 + x) (1 + x^{2}) (1 + x^{4}) . . . . (1 + x^{2 n})$ , then $\frac{d y}{d x}$ at x = 0 is

A

0

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B

-1

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C

1

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D

2

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Solution

The correct option is C
1

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

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

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

.......................................

= $(1 - x^{2 n}) (1 + x^{2 n})$ = $1 - x^{2 n + 1}$

$∴ y$ = $\frac{1 - x^{2 n + 1}}{(1 - x)}$

$∵ \frac{d y}{d x}$ = $\frac{(1 - x) (- 2 n + 1 . x^{2 n} - (1 - x^{2 n + 1}) (- 1)))}{(1 - x)^{2}}$

$∴ \frac{d y}{d x} {∣ ∣ ∣}_{x = 0}$ = $\frac{(1 - 0) (- 2^{n + 1} .0) - (1 - 0) (- 1)}{1}$

= 1

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Similar questions

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}) . . . . . . (1 + x^{2 n})

then the value of

(\begin{matrix} \frac{d y}{d x} \end{matrix})

x = 0

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

then the value of

{(\begin{matrix} \frac{d y}{d x} \end{matrix})}_{x = 0}

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

\frac{d y}{d x}

Q. Assertion :If

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

\frac{d y}{d x}

x = 0

1

y = \frac{1 - x^{2^{n + 1}}}{1 - x}

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