If y - 1 1 - x n - m - x p - m - 1 1 - x m - n - x p - n - 1 1 - x m - p - x n - p then d y d x at x - e m n p is equal to

The correct option is A 0 y=(11+(x^nx^m)+(x^px^m))+(11+(x^mx^n)+(x^px^n))+(11+(x^mx^p)+(x^nx^p)) =(x^mx^m+x^n+x^p)+(x^nx^n+x^m+x^p)+(x^px^p+x^m+ ...

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Basic Differentiation Rule

If y = 1 1 ...

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 $\frac{d y}{d x}$ at $x = e^{m^{n p}}$ is equal to

e^{m n p}

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e^{m n / p}

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e^{n p / m}

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Solution

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Similar questions

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{d y}{d x}

e^{m^{n^{p}}}

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}}, t h e n \frac{d y}{d x}

\sqrt[n]{x^{P}} = x^{m}

p

x^{m}

(1 + x)^{m} + 2 (1 + x)^{n - 1} + 3 (1 + x)^{n - 2} + (n - m + 1) (1 + x)^{m}

0 \geq m \geq n i s^{n + 2} C_{m + 2}

I_{m, n} = \int_{0}^{1} x^{m - 1} (1 - x)^{n - 1} d x,

m, n \geq 1

\int_{0}^{1} \frac{x^{m - 1} + x^{n - 1}}{(1 + x)^{m + n}} d x = α I_{m, n}, α \in R,

α

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