If y-1-1-xn-m-xp-m-1-1-xm-n-xp-n-1-1-xm-p-xn-p then dy-dx-

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Dividing a Monomial by a Monomial

If y=1/1+xn-m...

Question

If $y = \frac{1}{1 + x^{n - m} + x^{p - m}} + \frac{1}{1 + x^{m - n} + x^{p - n}} + \frac{1}{1 + x^{m - p} + x^{n - p}}$ then $\frac{d y}{d x} =$

$1$

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$- 1$

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$0$

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None of these

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Solution

The correct option is C
$0$

Explanation for the correct option.
Step 1: Simplify $y$
$\begin{array}{rcl} y & = & \frac{1}{1 + x^{n - m} + x^{p - m}} + \frac{1}{1 + x^{m - n} + x^{p - n}} + \frac{1}{1 + x^{m - p} + x^{n - p}} \\ = & \frac{1}{x^{- m} (x^{m} + x^{n} + x^{p})} + \frac{1}{x^{- n} (x^{n} + x^{m} + x^{p})} + \frac{1}{x^{- p} (x^{p} + x^{m} + x^{n})} \\ = & \frac{x^{m}}{x^{m} + x^{n} + x^{p}} + \frac{x^{n}}{x^{m} + x^{n} + x^{p}} + \frac{x^{p}}{x^{m} + x^{n} + x^{p}} \\ = & \frac{x^{m} + x^{n} + x^{p}}{x^{m} + x^{n} + x^{p}} \\ = & 1 \end{array}$
Step 2: Find $\frac{d y}{d x}$
Differentiating $y$ with respect to $x$ , we get
$\frac{d y}{d x} = 0$
Hence, option C is correct.

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(b)

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then

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Q. If

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

then

\frac{d y}{d x}

x = e^{m^{n p}}

is equal to

Q. If

x_{1} = 1

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, then

x_{n}

is equal to :

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If $y = \frac{1}{1 + x^{n - m} + x^{p - m}} + \frac{1}{1 + x^{m - n} + x^{p - n}} + \frac{1}{1 + x^{m - p} + x^{n - p}}$ then $\frac{d y}{d x} =$