document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1022092>Question 1022092</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi! Please help me with this question. \r<BR>\n" );
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document.write( "Differentiate  y= sqrtx / (x^2+1)  using the quotient rule \r<BR>\n" );
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document.write( "My work:\r<BR>\n" );
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document.write( "f'(x) =((x^2+1)(x^(1/2))' - (x^(1/2))(x^2+1)')/(x^2+1)^2\r<BR>\n" );
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document.write( "= ((x^2+1)(1/2x^(-1/2)) - (x^(1/2))(2))/ (x^2+1)^2\r<BR>\n" );
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document.write( "= (1/2x^(5/2) + 1/2x^(-1/2)-2x^(1/2))/ (x^2+1)^2\r<BR>\n" );
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document.write( "I can't seem to get the answer which is  (1-3x^2)/((2sqrtx)(x^2+1)^2)\r<BR>\n" );
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document.write( "Thank you! \r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #637804 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=637804\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Look at the numerator first<BR>\n" );
document.write( "it is (x^2+1)*(1/2 sqrt (x))- sqrt(x)*2x<BR>\n" );
document.write( "Now put this over a common denominator of 2 sqrt(x).  That means multiplying the second term by 2(sqrt(x)), and that will be 4x^2.<BR>\n" );
document.write( "The common denominator of 2 sqrt (x) goes into the denominator with (x^2+1)^2.<BR>\n" );
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document.write( "Now, the numerator is (x^2+1)-4x^2, the common denominator now part of the answer denominator.<BR>\n" );
document.write( "That is 1-3x^2. </SPAN>\n" );
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