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Thread: Proving the formula for the directional derivatives of the of the sum and dot product

  1. #1
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    Proving the formula for the directional derivatives of the of the sum and dot product

    Define the directional derivative of a function $\displaystyle \textbf{f}$ at $\displaystyle \textbf{c}$ in the direction $\displaystyle \textbf{u}$ by
    $\displaystyle \textbf{f}\hspace{0.04in}'(\textbf{c};\textbf{u}) = \lim_{h \rightarrow 0} \frac{\textbf{f}(\textbf{c}+h\textbf{u}) - f(\textbf{c})}{h},$
    whenever the limit on the right exists.

    Let $\displaystyle \textbf{f}$ and $\displaystyle \textbf{g}$ be functions with values in $\displaystyle \mathbb{\textbf{R}}^m$ such that the directional derivatives $\displaystyle \textbf{f}\hspace{0.04in}'(\textbf{c};\textbf{u})$ and $\displaystyle \textbf{g}'(\textbf{c};\textbf{u})$ exist.

    Prove that the sum $\displaystyle \textbf{f+g}$ and dot product $\displaystyle \textbf{f}$ $\displaystyle \bullet$ $\displaystyle \textbf{g}$ have directional derivatives given by
    $\displaystyle (\textbf{f+g})'(\textbf{c};\textbf{u}) = \textbf{f}\hspace{0.04in}'(\textbf{c};\textbf{u}) + \textbf{g}'(\textbf{c};\textbf{u})$
    and
    $\displaystyle (\textbf{f} \bullet \textbf{g})'(\textbf{c};\textbf{u)} = \textbf{f}(\textbf{c}) \bullet \textbf{g}'(\textbf{c};\textbf{u}) + \textbf{g}(\textbf{c}) \bullet \textbf{f}\hspace{0.04in}'(\textbf{c};\textbf{u}).$


    $\displaystyle \underline{\textbf{What I've Tried}}$
    I've figured out how to prove the sum part already but I'm having trouble with the dot product part. I tried to do it backwards

    $\displaystyle f(c) \bullet g'(c;u) + g(c) \bullet f'(c;u)$

    $\displaystyle = f(c) \bullet \lim_{h \rightarrow 0} \frac{g(c+hu)-g(c)}{h} + g(c) \bullet \lim_{h \rightarrow 0} \frac{f(c+hu)-f(c)}{h}$

    $\displaystyle = \lim_{h \rightarrow 0} f(c) \bullet \frac{g(c+hu)-g(c)}{h} + \lim_{h \rightarrow 0} g(c) \bullet \frac{f(c+hu)-f(c)}{h}$

    $\displaystyle =\lim_{h \rightarrow 0} \frac{f(c) \bullet g(c+hu) - f(c) \bullet g(c) + g(c) \bullet f(c+hu) - g(c)\bullet f(c)}{h}$

    $\displaystyle \textbf{But I'm not sure how that will arrive to}$:

    $\displaystyle = \lim_{h \rightarrow 0} \frac{f(c+hu)\bullet g(c+hu) - f(c) \bullet g(c)}{h}$

    $\displaystyle = \lim_{h \rightarrow 0} \frac{(f \bullet g)(c+hu) - (f \bullet g)(c)}{h}$

    $\displaystyle = (f \bullet g)'(c;u)$
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  2. #2
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    Re: Proving the formula for the directional derivatives of the of the sum and dot pro

    Hint: Look at proving the product rule for normal, single variable derivatives. There's a term you add and subtract in the numerator to make everything cancel out.
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