document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1034903>Question 1034903</A>: <I><SPAN STYLE=\"word-break: break-all;\"> sort of a calculus question... Find the derivative of the inverse function (f^-1)'(a) if f(x)=(e^x)+x and a=1. I've been stuck on this problem for over an hour now. I figured you should find the inverse before the derivative but I'm having trouble with that. I've tried taking the natural logarithm of both sides. I can't get x by itself. Even when I try to use other logarithm rules. I also asked my classmates and they are stuck too. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #649563 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=649563\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the derivative of e^x + x is e^x + 1.<BR>\n" );
document.write( "i'm not that smart.<BR>\n" );
document.write( "i used an online derivative calculator to find it.\r<BR>\n" );
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document.write( "you get y = e^x + 1 is the equation of the derivative of y = e^x + x.\r<BR>\n" );
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document.write( "the inverse y = e^x + 1 is x = e^y + 1.<BR>\n" );
document.write( "you interchange the variables.<BR>\n" );
document.write( "x = y and y = x<BR>\n" );
document.write( "that's your inverse equation.\r<BR>\n" );
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document.write( "subtract 1 from both sides of that equation to get x - 1 = e^y<BR>\n" );
document.write( "take the natural log of both sides of that equation to get ln(x-1) = ln(e^y)<BR>\n" );
document.write( "since ln(e^y) = y*ln(e), and since ln(e) = 1, the equation becomes:<BR>\n" );
document.write( "ln(x-1) = y<BR>\n" );
document.write( "this can be shown as y = ln(x-1).\r<BR>\n" );
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document.write( "y = ln(x-1) is equivalent to x = e^y + 1 and is therefore the derivative equation of y = e^x + x, as far as i can tell.\r<BR>\n" );
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document.write( "what i end up with is that the inverse equation of the derivative of y = e^x + x winds up being y = ln(x-1).\r<BR>\n" );
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document.write( "if you're saying that f(x) = ln(x-1) is the derivative equation, and you want to find f(a), then the equation becomes f(a) = ln(a-1).\r<BR>\n" );
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document.write( "if a = 1, then the equation becomes f(1) = ln(1-1) = ln(0) which is undefined.\r<BR>\n" );
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document.write( "i'm not totally sure i did the right thing, but it kind of makes sense to me.<BR>\n" );
document.write( "don't take it as gospel though, because i'm not totally sure it's what you want.<BR>\n" );
document.write( "it's the best i can do however, so take it for what it's worth.\r<BR>\n" );
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