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Explore properties of an alternative definition of the derivative

Consider the following different definition of the a derivative:

    \[ D^* f(x) = \lim_{h \to 0} \frac{f^2(x+h) - f^2(x)}{h}, \]

where f^2(x) means [f(x)]^2.

  1. Find formulas for the “derivative” of the sum, difference, product, and quotient of functions.
  2. Find an expression for D^* f(x) in terms of the usual derivative D f(x).
  3. Find the functions f for which D^* f = Df.

  1. Sum. Using the formula in the alternative definition,

        \begin{align*}  D^* &(f(x) + g(x)) = \lim_{h \to 0} \frac{(f(x+h) + g(x+h))^2 - (f(x) + g(x))^2}{h} \\[8pt]  &= \lim_{h \to 0} \frac{f^2(x+h) + 2f(x+h)g(x+h) + g^2 (x+h) - f^2(x) - 2f(x)g(x) - g^2(x)}{h} \\[8pt]  &= \lim_{h \to 0} \left( \frac{f^2(x+h) - f^2(x)}{h} \right) + 2 \cdot \lim_{h \to 0} \left( \frac{f(x+h)g(x+h) - f(x)g(x)}{h} \right) \\[8pt]  &\qquad + \lim_{h \to 0} \left( \frac{g^2(x+h) - g^2(x)}{h} \right) \\[8pt]  &= D^* f(x) + D^* g(x) + 2 \cdot \lim_{h \to 0} \left( \frac{f(x+h)g(x+h) - f(x)g(x)}{h} \right). \end{align*}

    Now, for the remaining limit term we multiply by \frac{f(x+h)g(x+h) + f(x)g(x)}{f(x+h)g(x+h) + f(x)g(x)} (thanks to abcd in the comments for the suggestion!). So, we then have

        \begin{align*}  &= D^* f(x) + D^* g(x) + 2 \cdot \lim_{h \to 0} \left( \frac{f(x+h)g(x+h) - f(x)g(x)}{h} \cdot \frac{f(x+h)g(x+h) + f(x)g(x)}{f(x+h)g(x+h) + f(x)g(x)} \right) \\[9pt]  &= D^* f(x) + D^* g(x) + 2 \cdot \lim_{h \to 0} \left( \frac{f^2(x+h)g^2(x+h) - f^2(x) g^2(x)}{h \cdot (f(x+h)g(x+h) + f(x)g(x))} \right) \\[9pt]  &= D^* f(x) + D^* g(x) + \lim_{h \to 0} \left( \frac{2}{f(x+h)g(x+h) + f(x)g(x)} \cdot \frac{f^2(x+h)g^2(x+h) - f^2(x)g^2(x)}{h} \right) \\[9pt]  &= D^* f(x) + D^* g(x) + \frac{1}{f(x)g(x)} \cdot \lim_{h \to 0} \frac{(f(x+h)g(x+h))^2 - (f(x)g(x))^2}{h}. \end{align*}

    Now, for the remaining limit, we use the product rule for the derivative in this alternative definition, which we derive below. (This last limit is D^* (f(x)g(x)). Below we derive that D^* (f(x)g(x)) = g^2 D^* f + f^2 D^* g.) Therefore, we have

        \begin{align*}  D^* (f(x)+g(x)) &= D^* f(x) + D^* g(x) + \frac{1}{f(x) g(x)} \cdot \lim_{h \to 0} \frac{(f(x+h)g(x+h))^2 - (f(x)g(x))^2}{h} \\[9pt]  &= D^* f(x) + D^* g(x) + \frac{1}{f(x) g(x)} \cdot \left( g^2 (x) D^* f(x) + f^2(x) D^* g(x) \right) \\[9pt]  &= D^* f(x) + D^* g(x) + \frac{g(x)}{f(x)} D^* f(x) + \frac{f(x)}{g(x)} D^* g(x) \\[9pt]  &= \left( 1 + \frac{g(x)}{f(x)} \right)D^* f(x) + \left( 1 + \frac{f(x)}{g(x)} \right) D^* g(x). \end{align*}

    Difference. This follows just as the above derivation for D^* (f(x) + g(x)) except that we get minus signs:

        \[ D^* (f(x) - g(x)) = \left( 1 - \frac{g}{f} \right) D^* f + \left( 1 - \frac{f}{g} \right) D^* g. \]

    Product. Using the alternative definition of the “derivative” we compute,

        \begin{align*}  D^* &(f(x)g(x)) = \lim_{h \to 0} \frac{(f(x+h)g(x+h))^2 - (f(x)g(x))^2}{h} \\[8pt]  & = \lim_{h \to 0} \frac{f^2(x+h) g^2(x+h) - f^2(x)g^2(x)}{h} \\[8pt]  & = \lim_{h \to 0} \frac{f^2(x+h) g^2(x+h) - f^2(x) g^2(x) + f^2(x+h)g^2(x) - f^2(x+h)g^2(x)}{h}  \intertext{(where we added and subtracted $f^2(x+h)g^2(x)$ in the numerator)}  & = \lim_{h \to 0} \left(g^2(x) \frac{f^2(x+h) - f^2(x)}{h} \right) + \lim_{h \to 0} \left(f^2(x+h) \frac{g^2(x+h) - g^2(x)}{h}\right) \\[8pt]  & = g^2(x) D^*f + f^2(x) D^*g. \end{align*}

    Quotient. Again, we start with the alternative definition and compute,

        \begin{align*}  D^* & \left( \frac{f(x)}{g(x)} \right) = \lim_{h \to 0} \frac{ \frac{f^2(x+h)}{g^2(x+h)} - \frac{f^2(x)}{g^2(x)}}{h} \\[8pt]  &= \lim_{h \to 0} \left( \frac{g^2(x) f^2(x+h) - f^2(x) g^2(x+h)}{h \cdot g^2(x+h) g^2(x)} \right) \\[8pt]  &= \frac{1}{g^4(x)} \lim_{h \to 0} \left( \frac{g^2(x) f^2(x+h) - f^2(x) g^2(x+h) + g^2(x)f^2(x) - g^2(x) f^2(x)}{h}\right) \\[8pt] \intertext{(since $\lim_{h \to 0} g(x+h) = g(x)$ we had a $g^4(x)$ term in the denominator that we pulled out of the limit since $g(x)$ is a constant in $h$.  Further, we added and subtracted $g^2(x)f^2(x)$ in the numerator.)}  &= \frac{1}{g^4 (x)} \lim_{h \to 0} \left( g^2 (x) \frac{f^2(x+h) - f^2(x)}{h} - f^2(x) \frac{g^2(x+h) - g^2(x)}{h} \right) \\[8pt]  &= \frac{g^2 D^* f - f^2 D^* g}{g^4}.  \end{align*}

  2. Expressing D^* f in terms of the usual derivative Df we have,

        \begin{align*}  D^* f(x) &= \lim_{h \to 0} \frac{f^2(x+h) - f^2(x)}{h} \\[8pt]   &= \lim_{h \to 0} \frac{(f(x+h) + f(x))(f(x+h) - f(x))}{h} \\[8pt]  &= 2 f(x) \cdot Df(x).  \end{align*}

    We used the same idea we used in deriving some of the above formulas that \lim_{h \to 0} f(x+h) = f(x), and that f(x) is a constant in h, so we can pull it out of the limit.

  3. Now, to find the function f such that Df = D^* f we set the two expressions equal and solve for f (using D^* f = 2f Df from part (c)),

        \[ Df = D^* f \quad \implies \quad Df = 2f(x) Df \quad \implies \quad Df = 0 \text{ or } 2f(x) = 1. \]

    In either case we have f(x) = c for some constant c.


  1. Hteica says:

    Hi, in (a), I don’t think we are allowed to multiply [ (fg)(x+h) + (fg)(x) ] / [ (fg)(x+h) + (fg)(x) ], what happens when f is defined to be -1 for all values except at x where f(x)=1, and g defined as constant 1 for any values (but the limit of D*f and D*g do exist), we get (fg)(x+h) + (fg)(x)= -1+1=0, which means we were multiplying by 0/0 all along, which is not ok. Thanks!

  2. abcd says:

    Also, I believe that your solution for part a (formula for D^*(f+g)) is wrong. In your solution, you reduced \lim_{h \to 0} \left( \frac{(fg)(x+h) - (fg)(x)}{h} \right) to two different expressions: (f/g)D^*g, and (g/f)D^*f.

    If both are true, we can get a different expression for D^*(f+g), e.g. (1+2f/g)D^*g + D^*f which is clearly different from the answer.

    I believe the error was made when you cancelled f(x+h)/f(x) and similar terms in the fraction because for the limit \lim(A+B) it is illegal to cancel a term in only A (or only B) if B (or A) diverges.

    • RoRi says:

      Hey, thanks for this and the other comment. You’re right, my solution contains the error you mention. Fixing it now.

  3. abcd says:

    For part a, I think that we can deal with the 2 [ (fg)(x+h) - (fg)(x)]/h term by multiplying by [ (fg)(x+h) + (fg)(x) ] / [ (fg)(x+h) + (fg)(x) ] to give

        \[ \frac{2}{(fg)(x+h) + (fg)(x)} \cdot  \frac{(fg)^2 (x+h) - (fg)^2 (x)}{h} \]

    which reduces to D^*(fg)/fg.

    Applying the product rule derived later reduces the quotient, thus skipping the painful manipulation.

    • Awamoki says:

      abcd, there is another way to prove the part a without applying the product rule. We can add and subtract (f(x)^2)(g(x+h)^2) at the numerator after you multiplied by [ (fg)(x+h) + (fg)(x) ] / [ (fg)(x+h) + (fg)(x) ] and rearrange the sum.

  4. tasos says:

    Alternatively for part a, D^* f=(f^2)' \ \implies \  (f^2)' is continuous and from composition f is differentiable and continuous hence we can get D^*f =2f f' from part b (where apostol implies differentiability of f) to make the calculations much more painless. The direct results are recursive formulas of normal derivatives rather than D^* f though.

    • RoRi says:

      Yes, that works also and is decidedly less painful. I figured since Apostol put it before part (b) that he wanted us to suffer and compute it out the long way by hand to see how it works. Maybe I overestimate how mean he is though.

      PS. I hope you don’t mind that I edit LaTeX into your comments. Let me know if it’s a problem or if I introduce any errors.

      • tasos says:

        I’m sure that’s what he had in mind. And no, I honestly appreciate you introducing latex into my comments.

      • Mihajlo says:

        It doesn’t seem to be the matter of meanness, but that you can learn more if you do a) without using the result of b).

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