Question

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

(a) 0, n!, mPn x^(m-n)

(b) mPn x^(m-n), n!, 0

(c) 0, n!, mCn x^(m-n)

(d) mCn x^(m-n), n!, 0

The question was asked in exam.

I'm obligated to ask this question of The nth Derivative of Some Elementary Functions in division Differential Calculus of Engineering Mathematics

Answer 1

The correct choice is (a) 0, n!, mPn x^(m-n)

For explanation I would say: For, m > n

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

Since m>n, m-n=0 hence this cycle will moves upto (m-n) times and at last

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

Hence,

\(\frac{d^n (x^m)}{dx^n}=m_{P_n} x^{(m-n)}\) ………. (1)

For m=n, from equation 1,

\(\frac{d^n (x^n)}{dx^n}=n_{P_n} x^{(n-n)}=n!\)

From m<n, from equation 1,

\(\frac{d^n (x^m)}{dx^n}=m_{P_n} x^{(m-n)}=0\) (Because, \(m_{P_n}=0\) for m<n)