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Question

Using mathematical induction, prove that ddx(xn)=nxn1 for all positive integers n.

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Solution

Check the given expression for n=1
L.H.S. =ddx(x1)=dxdx=1
R.H.S. =1.x11=1
Hence, it is true for n=1

Let the expression be true for n=k
ddx(xk)=kxk1
Now, check for n=k+1
L.H.S. =ddx(xk+1)=ddx(xk.x)=xddx(xk)+xkddx(x)=x(kxk1)+xk(1)=kxk+xk=(k+1)xk=(k+1)x(k+1)1

Hence, the given expression is true for n=k+1 too.
By principle of mathematical induction, ddx(xn)=nxn1

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