Q19  Using mathematical induction prove that \frac{d}{dx} (x^n) = nx ^{n-1} for all positive  integers n.

Answers (1)

Given equation is
\frac{d}{dx} (x^n) = nx ^{n-1}
We need to show that    \frac{d}{dx} (x^n) = nx ^{n-1} for all positive  integers n
Now,
For ( n = 1)    \Rightarrow \frac{d(x^{1})}{dx}= 1.x^{1-1}= 1.x^0=1
Hence, true for n = 1
For (n = k)     \Rightarrow \frac{d(x^{k})}{dx}= k.x^{k-1}
Hence, true for n = k
For ( n = k+1) \Rightarrow \frac{d(x^{k+1})}{dx}= \frac{d(x.x^k)}{dx}
                                                = \frac{d(x)}{dx}.x^k+x.\frac{d(x^k)}{dx}
                                                = 1.x^k+x.(k.x^{k-1}) = x^k+k.x^k= (k+1)x^k
Hence, (n = k+1) is true whenever (n = k) is true
Therefore, by the principle of  mathematical induction  we can say that  \frac{d}{dx} (x^n) = nx ^{n-1} is true for  all positive  integers n

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