Question

(a) If \(F\left( x \right) = f\left( x \right)g\left( x \right)\), where fand ghave derivative of all orders, show that \(F'' = f''g + {\bf{2}}f'g' + fg''\).

(b) Find the similar formulas for \(F'''\), and \({F^{\left( {\bf{4}} \right)}}\).

(c) Guess a formula for \({F^{\left( n \right)}}\).

Short answer

(a)The equation \(F'' = f''g + 2f'g' + fg''\) is true.

(b)The value of \(F''' = f'''g + 3f''g' + 3f'g'' + fg'''\) and \({F^{\left( 4 \right)}} = {f^{\left( 4 \right)}}g + 4f'''g' + 6f''g'' + 4f'g''' + f{g^{\left( 4 \right)}}\)

(c) \({F^{\left( n \right)}} = {f^{\left( n \right)}}g + n{f^{\left( {n - 1} \right)}}g' + \left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right){f^{\left( {n - 2} \right)}}g'' + ... + \left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right){f^{\left( {n - k} \right)}}{g^{\left( k \right)}} + .... + nf'{g^{\left( {n - 1} \right)}} + f{g^{\left( n \right)}}\)

Step-by-step solution

(a) Step 1: Find the value of \(F''\)

The value of \(F'\) is calculated using the Product Rule.

\(\begin{aligned}F' &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {fg} \right)\\ &= fg' + f'g\end{aligned}\)

Differentiatethe equation again to find \(F''\).

\(\begin{aligned}F'' &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {fg' + f'g} \right)\\ &= fg'' + g'f' + f'g' + gf''\\ &= f''g + 2f'g' + fg''\end{aligned}\)

So, the value of \(F''\) is \(f''g + 2f'g' + fg''\).

(b) Step 2: Find the value of \(F'''\) and \({F^{\left( {\bf{4}} \right)}}\) 

Differentiate the equation \(F'' = f''g + 2f'g' + fg''\).

\(\begin{aligned}F''' &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {f''g + 2f'g' + fg''} \right)\\ &= f''g' + gf''' + 2f'g'' + 2g'f'' + fg''' + g''f'\\ &= f'''g + 3f''g' + 3f'g'' + fg'''\end{aligned}\)

Differentiate the equation \(F''' = f'''g + 3f''g' + 3f'g'' + fg'''\).

\(\begin{aligned}{F^{\left( 4 \right)}} &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {f'''g + 3f''g' + 3f'g'' + fg'''} \right)\\ &= {f^{\left( 4 \right)}}g + f'''g' + 3f'''g' + 3f''g'' + 3f''g'' + 3f'g''' + f'g''' + f{g^{\left( 4 \right)}}\\ &= {f^{\left( 4 \right)}}g + 4f'''g' + 6f''g'' + 4f'g''' + f{g^{\left( 4 \right)}}\end{aligned}\)

So, the value of \(F''' = f'''g + 3f''g' + 3f'g'' + fg'''\), and \({F^{\left( 4 \right)}} = {f^{\left( 4 \right)}}g + 4f'''g' + 6f''g'' + 4f'g''' + f{g^{\left( 4 \right)}}\).

(c) Step 3: Guess the formula for \({F^{\left( n \right)}}\)

Using the Binomial Theorem, the guess can be made for \({F^{\left( n \right)}}\).

\({F^{\left( n \right)}} = {f^{\left( n \right)}}g + n{f^{\left( {n - 1} \right)}}g' + \left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right){f^{\left( {n - 2} \right)}}g'' + ... + \left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right){f^{\left( {n - k} \right)}}{g^{\left( k \right)}} + .... + nf'{g^{\left( {n - 1} \right)}} + f{g^{\left( n \right)}}\)

The expression \(\left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right)\) represents \(\frac{{n!}}{{k!\left( {n - k} \right)!}}\).

So, the expression for \({F^{\left( n \right)}}\) is \({f^{\left( n \right)}}g + n{f^{\left( {n - 1} \right)}}g' + \left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right){f^{\left( {n - 2} \right)}}g'' + ... + \left( {\begin{array}{*{20}{c}}n\\k\end{array}} \right){f^{\left( {n - k} \right)}}{g^{\left( k \right)}} + .... + nf'{g^{\left( {n - 1} \right)}} + f{g^{\left( n \right)}}\).