Any help you can give me would be appreciated!(adsbygoogle = window.adsbygoogle || []).push({});

The teacher wrote the solution to the problem on the board without giving much explanation of how he got there, and now he wants the third derivative, so I was wondering if you could help answer how he got the second.

g(x) [the lim. as x goes to zero] = (e^x - 1) / x , where x can not equal zero

then he wrote

1, x = 0 I'm guessing this means to use the value 1 when x = 0

Then he wrote the first derivative of g(x) is

( xe^x - e^x + 1) / (x^2) , where x does not equal zero

and

1/2 x = 0

then the second derivative of g(x) :

g(x) [lim. as x goes to zero] ( ( (xe^x -e^x + 1) / x^2 ) - (1/2) ) / x

then goes to:

( 2xe^x - 2e^x + 2 - x^2 ) / 2x^3

then goes to:

( 2e^x + 2xe^x - 2e^x - 2x) / 6x^2 [the 2e^x's cancel]

this goes to:

( e^x - 1 ) / 3x = 1/3

Now I have to do the third derivative. If you could help me understand this second derivative it may help me solve the third. Thanks in advance for any time you spend on this.

-Kevin

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# L'Hopital / Second Derivative

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