Question

Prove Formula 3 of Theorem 5.

Short answer

The proof of the statement \(\frac{d}{{dt}}(f(t)u(t)) = f'(t)u(t) + f(t)u'(t)\) is explained.

Step-by-step solution

Step 1: The statement to prove

We have to prove \(\frac{d}{{dt}}(f(t)u(t)) = f'(t)u(t) + f(t)u'(t)\)

\(\begin{array}{l}let\;\;\;\;\;\;\;\;\;\;\;u(t) = \left\langle {{f_1}(t),{f_2}(t),{f_3}(t)} \right\rangle \\then\;\;\;f(t)u(t) = \left\langle {f(t){f_1}(t),f(t){f_2}(t),f(t){f_3}(t)} \right\rangle \end{array}\)

Since f(t) is scalar function. We now have:

\(\begin{array}{l}\frac{d}{{dt}}(f(t)u(t)) = \frac{d}{{dt}}\left\langle {f(t){f_1}(t),f(t){f_2}(t),f(t){f_3}(t)} \right\rangle \\ = \left\langle {\frac{d}{{dt}}f(t){f_1}(t),\frac{d}{{dt}}f(t){f_2}(t),\frac{d}{{dt}}f(t){f_3}(t)} \right\rangle \end{array}\)

Step 2: Further simplification

\(\begin{array}{l} = \left\langle {f'(t){f_1}(t) + f(t){f_1}'(t),f'(t){f_2}(t) + f(t){f_2}'(t),f'(t){f_3}(t) + f(t){f_3}'(t)} \right\rangle \\ = \left\langle {f'(t){f_1}(t),f'(t){f_2}(t),f'(t){f_3}(t)} \right\rangle + \left\langle {f(t){f_1}'(t) + f(t){f_2}'(t) + f(t){f_3}'(t)} \right\rangle \\ = f'(t)\left\langle {{f_1}(t),{f_2}(t),{f_3}(t)} \right\rangle + f(t)\left\langle {{f_1}'(t) + {f_2}'(t) + {f_3}'(t)} \right\rangle \\ = f'(t)u(t) + f(t)u'(t)\end{array}\)

The proof of the statement \(\frac{d}{{dt}}(f(t)u(t)) = f'(t)u(t) + f(t)u'(t)\) is explained.