What is the value of \(\frac{{{d^n}({x^m})}}{{d{x^n}}}\) for m < n, m = n, m > n?

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BPSC Assistant Professor Mechanical 2022 Official Paper
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  1. 0, n!, mPx(m-n)
  2. mPx(m-n), n!, 0
  3. 0, n!, mCx(m-n)
  4. mCn x(m-n), n!, 0

Answer (Detailed Solution Below)

Option 1 : 0, n!, mPx(m-n)

Detailed Solution

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Explanation:

\(\frac{{{d^n}({x^m})}}{{d{x^n}}}\) = ? ; for m < n, m = n, and m > n

∵ \(\frac{{{d^n}({x^m})}}{{d{x^n}}}\) = \(m\frac{d^{n~-~1 ~}(x^{m~-~1})}{dx^{n~-~1}}~=~m(m~-~1)\frac{d^{n~-~2 ~}(x^{m~-~2})}{dx^{n~-~2}}~=~........\)

Case: I

Since m > n, hence the cycle will move up to (m - n) times, and at last

\(\frac{{{d^n}({x^m})}}{{d{x^n}}}\) = \(m(m~-~1)(m~-~2)...................[m~-~(n~-~1)]x^{m~-~n}\)

Hence, for m > n,

\(\frac{{{d^n}({x^m})}}{{d{x^n}}}\) = mPx(m - n)........................................................(1)

Case: II

For m = n:

From equation (1),

\(\frac{{{d^n}({x^m})}}{{d{x^n}}}\) = nPx(n - n) =  n!

Case: III

For m < n:

\(\frac{{{d^n}({x^m})}}{{d{x^n}}}\) = mPx(m - n) = 0 ( Because mP= 0; For m < n )

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