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Differentiating f(-x)?

  1. Apr 7, 2013 #1
    I came across this simple expression while doing some maths.

    If [tex]\frac{d}{dx}f(x)=g(x)[/tex]

    Then [tex]\frac{d}{dx}f(-x)=-g(-x)[/tex]



    Is this correct? How do we prove it?
     
  2. jcsd
  3. Apr 7, 2013 #2

    phyzguy

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    Do you know the chain rule? This is a simple example of it.
     
  4. Apr 7, 2013 #3
    What are the specifics?

    :blushing:
     
  5. Apr 7, 2013 #4
    Chain rule: [itex] \frac{d[f(g(x))]}{dx} = \frac{d[f(g(x))]}{d[g(x)]} \frac{d[g(x)]}{dx} [/itex]. In your case, g(x) = -x.
     
  6. Apr 7, 2013 #5

    jbunniii

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    If you prefer a direct proof without using the chain rule, use the definition of the derivative. We can prove something slightly more general. If
    $$g(x) = \frac{d}{dx}f(x) = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}$$
    and ##a(x) = f(cx)## where ##c## is any nonzero real number,
    $$\begin{align}
    \frac{d}{dx} a(x) &= \lim_{h \to 0} \frac{a(x+h) - a(x)}{h} \\
    &= \lim_{h \to 0} \frac{f(c(x+h)) - f(cx)}{h} \\
    \end{align}$$
    Letting ##k = ch##, we note that ##h \to 0## if and only if ##k \to 0##, so the above is equivalent to
    $$\begin{align}
    \frac{d}{dx} a(x) &= \lim_{k \to 0} \frac{f(cx + k) - f(cx)}{k/c} \\
    &= c \lim_{k \to 0} \frac{f(cx + k) - f(cx)}{k} \\
    &= c g(cx) \\
    \end{align}$$
    Your case follows by setting ##c = -1##.
     
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