Question

If $\mathit{f}\left(x\right)\mathbf{\in }\mathit{F}\left(x\right)$and n is a positive integer prove that the of $\mathit{f}{\left(x\right)}^{\mathbf{n}}$ is $\mathit{n}\mathit{f}{\left(x\right)}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathit{f}\left(x\right)$.

Short answer

It is proved that $f{\left(x\right)}^n$ is .$nf{\left(x\right)}^{n-1}f\left(x\right)$

Step-by-step solution

Definition of polynomial

A polynomial $p\left(x\right)$ over a given field K is separable if its root are distinct in an algebraic closure of K . That is the number of root is equal to the degree of the polynomial.

Showing that  nfxn-1fx 

Let $f\left(x\right)\in F\left[x\right]$ and n is positive integer. To prove above claim put $n=1$as:

$\begin{array}{rcl}{\left[f{\left(x\right)}^1\right]}^1& =& 1·f{\left(x\right)}^{1-1}{f}^1\left(x\right)\\ & =& f{\left(x\right)}^0{f}^1\left(x\right)\\ & =& f^1\left(x\right)\\ & & \end{array}$

Thus the result holds for n=1. Assume the same result hold for n=k. That is:

${\left[f{\left(x\right)}^k\right]}^1=kf{\left(x\right)}^{k-1}{f}^1\left(x\right)\left(1\right)$

For $n=k+1$:

${\left[f{\left(x\right)}^{k+1}\right]}^1=\left(k+1\right)f{\left(x\right)}^k{f}^1\left(x\right)$

Consider the left hand side of the above equation ${\left[f{\left(x\right)}^{k+1}\right]}^1$.

Apply product rule for derivative :

$\begin{array}{rcl}{\left[f{\left(x\right)}^{k+1}\right]}^1& =& {\left(f{\left(x\right)}^k{f}^1\left(x\right)\right)}^1\\ & =& f{\left(x\right)}^k{f}^1\left(x\right)+{\left[f{\left(x\right)}^k\right]}^{1}f\left(x\right)\\ & & \end{array}$

From 1,

$\begin{array}{rcl}{\left[f{\left(x\right)}^{k+1}\right]}^1& =& f{\left(x\right)}^k{f}^1\left(x\right)+kf{\left(x\right)}^{k-1}{f}^1\left(x\right)f\left(x\right)\\ & =& f^1\left(x\right)\left(f{\left(x\right)}^k+kf{\left(x\right)}^{k-1}f\left(x\right)\right)\\ & =& f^1\left(x\right)\left(f{\left(x\right)}^k+kf{\left(x\right)}^k\right)\\ & =& f^1\left(x\right)f{\left(x\right)}^k\left(1+k\right)\\ & & \end{array}$

This gives:

${\left[f{\left(x\right)}^{k+1}\right]}^1=f^1\left(x\right)f{\left(x\right)}^k\left(1+k\right)$

Thus the result is true for $n=k+1$. Therefore,

${\left[f{\left(x\right)}^n\right]}^1=nf{\left(x\right)}^{n-1}f\left(x\right)$

It is proved that $f{\left(x\right)}^n$ is $nf{\left(x\right)}^{n-1}f\left(x\right)$.