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Question

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

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Solution

Let P(n)=ddx(xn)=nxn1Now,P(1) is ddx(x1)=1.x11 e.e.,ddx(x)=1, which is true. P(1) is true. Let P(m) be true, mϵ N ddx(xm)=mxm1 ddx(xm+1)=ddx{xmx}=xmddx(x)+x ddx (xm) (by product rule) =xm.1+x(mxm1)=(m+1)xm [using Eq. (i)] ddx(xm+1)={m+1}x(m+1)1 P(m+1) is true. Thus, P(1) is true and P(m) P(m+1), mϵ N.Hence, by induction P(n) is true for all nϵN


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