Question

Extended product rule: The product rule can be extended to the product of three functions.

1. Use the product rule twice to prove that if f, g, and h are differentiable, then \(\left( {fgh} \right)' = f'gh + fg'h + fgh'\).

1. Taking \(f = g = h\) in part (a), show that

\[\frac{{\bf{d}}}{{{\bf{d}}x}}{\left[ {f\left( x \right)} \right]^{\bf{3}}} = {\bf{3}}{\left[ {f\left( x \right)} \right]^{\bf{2}}}f'\left( x \right)\]

1. Use part (b) to differentiate \[y = {e^{{\bf{3}}x}}\].

Short answer

(a) The identity \(\left( {fgh} \right)' = f'gh + fg'h + fgh'\) is true.

(b) The identity \(\frac{{\rm{d}}}{{{\rm{d}}x}}{\left[ {f\left( x \right)} \right]^3} = 3{\left[ {f\left( x \right)} \right]^2}f'\left( x \right)\)is true.

(c) The derivative is \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = 3{e^{3x}}\).

Step-by-step solution

(a) Step 1: Apply the Product rule to the expression \(\left( {fgh} \right)'\)

Apply the product rule to find the derivative of \(\left( {fgh} \right)\).

\(\begin{array}{c}\left( {fgh} \right)' = \frac{{\rm{d}}}{{{\rm{d}}x}}\left[ {\left( {fg} \right)h} \right]\\ = \left( {fg} \right)h' + h\left( {fg} \right)'\\ = fgh' + h\left[ {fg' + f'g} \right]\\ = fgh' + fg'h + f'gh\end{array}\)

Hence, \(\left( {fgh} \right)' = fgh' + fg'h + f'gh\) is true.

(b)Step 2: Find an answer for part (b)

Use \(f = g = h\) for the equation \(\left( {fgh} \right)' = fgh' + fg'h + f'gh\).

\(\begin{array}{c}\left[ {{{\left\{ {f\left( x \right)} \right\}}^3}} \right]' = f'ff + ff'f + fff'\\\frac{{\rm{d}}}{{{\rm{d}}x}}{\left[ {f\left( x \right)} \right]^3} = 3fff'\\ = 3{\left[ {f\left( x \right)} \right]^2}f'\left( x \right)\end{array}\)

Hence, the identity \(\frac{{\rm{d}}}{{{\rm{d}}x}}{\left[ {f\left( x \right)} \right]^3} = 3{\left[ {f\left( x \right)} \right]^2}f'\left( x \right)\) is true.

(c)Step 3: Differentiate the function \(y = {e^{{\bf{3}}x}}\)

Differentiate the equation \(y = {e^{3x}}\).

\(\begin{array}{c}\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{e^{3x}}} \right)\\ = 3{e^{3x}}\end{array}\)

So, the required derivative is \(3{e^{3x}}\).