Using mathematical induction prove that ddx(xn)=nxn−1 for all positive integers n.
Let P(n)=ddx(xn)=nxn−1Now,P(1) is ddx(x1)=1.x1−1 e.e.,ddx(x)=1, which is true.∴ P(1) is true. Let P(m) be true, mϵ N⇒ ddx(xm)=mxm−1⇒ ddx(xm+1)=ddx{xmx}=xmddx(x)+x ddx (xm) (by product rule) =xm.1+x(mxm−1)=(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