Prove by induction that `{prod_(r=0)^(n)f_(r)(x)}'=sum_(i=1)^(n){f_1(x)f_2(x)....f_(i)'(x)....f_(n)(x)}`,
where dash denotes derivative with respect to x.

Explanation

Let `P(n):{prod_(r=0)^(n)f_(r)(x)}^(')=sum_(i=1)^(n){f_(1)(x)f_(2)(x).....f_(1)(x)....f_(n)(x)}`
Step I For ` n =1` ,
LHS of Eq. (i) `={prod_(r=1)^(1)f_(r)f(x)}^(')={f_(1)(x)}^(')=f_(1)^(')(x)`
RHS of Eq. (i) `=sum_(i=1)^(1){f_1(x)f_(2)(x)...f_(1)^(')(x)... f_(1)(x)}`
which is true for `n=1`.
Step II Assume it is true for `n=k` , then
`P(k):{prod_(r=1)^(k)f_(r)(x)}^(')=sum_(i=1)^(k){f_1(x)f_2(x).....f_1(x)....f_k(x)}`
Step III For `n=k+1`,
LHS`={prod_(r=1)^((k+1))f_r(x)}={prod_(r=1)^(k)f_r(x).f_(k+1)(x)}^(')`
`=prod_(r=1)^(k)f_r(x).f_(k+1)^(')(x)+f_(k+1)(x){prod_(r=1)^(k)f_r(x)}^'`
`= prod_(r=1)^(k)f_r(x).f_(k+1)^(')(x)+f_(k+1)(x).sum_(i=1)^(k){f_1(x).f_2(x).....f_(k+1)^(')....f_(k)(x)}`
`={f_1(x)f_2(x)....f_k(x)}f_(k+1)^(')(x)+f_(k+1)(x) sum_(i=1)^(k){f_1(x)f_2(x)....f_(i) '(x).....f_(k)(x)}`
`= sum_(i=1)^(k+1){f_19x)f_2(x).....f_(i)'(x)....f_(k+1)(x)}=RHS`
This shows that the result is true for `n=k+1`. Hence , by the principle of mathematical induction , the result is true for all `n in N`.