Derivative of e − x

Question

Derivative of       e          −      x

Layne Bailey · Accepted Answer

Note that       e          −      x       is a composition: we are composing the function g(x)=−x with the function   f  (  u  )  =      e    u  . By the Chain Rule, we have:      d          d      x            e          −      x        =      d          d      x        f            (        g  (  x  )            )        =      f    ′    (  g  (  x  )  )      g    ′    (  x  )  .Now, g(x)=−x, so g′(x)=−1. And   f  (  u  )  =      e    u  , so       f    ′    (  u  )  =      e    u  . Hence      d          d      x            e          −      x        =      f    ′    (  g  (  x  )  )      g    ′    (  x  )  =      e          g      (      x      )        (  −  1  )  =      e          −      x        (  −  1  )  =  −      e          −      x        .Therefore,      d          d      x        (  4      e    x    +  3      e          −      x        )  =  4      d          d      x            e    x    +  3      d          d      x            e          −      x        =  4      e    x    +  3  (  −      e          −      x        )  =  4      e    x    −  3      e          −      x        .

William Santiago · Accepted Answer

Firstly, we can use chain rule, which is:

\frac{d}{d x} f (g (x)) = f^{'} (g (x)) g^{'} (x)

where

f (x) = e^{x}

and g(x)=−x (so that

f (g (x)) = e^{- x}

). We know that

\frac{d}{d x} e^{x} = e^{x}

and

\frac{d}{d x} - x = - 1

so we get the derivative:

\frac{d}{d x} e^{- x} = e^{- x} * - 1 = - e^{- x}

To know why this occurs, we use first principles:

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

so thus

\frac{d}{d x} e^{- x} = lim_{h \to 0} \frac{e^{- (x + h)} - e^{- x}}{h} = lim_{h \to 0} e^{- x} (\frac{e^{- h} - 1}{h})

now, from the (a?) definition of e:

lim_{h \to 0} \frac{e^{- h} - 1}{h} = - 1

and thus

\frac{d}{d x} e^{- x} = - e^{- x}

Derivative of e

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