document.write( "Question 418868: Find an equation of the tangent line at the point indicated.
\n" ); document.write( "f (x) = log6 (5x + x^-5), x = 1 \n" ); document.write( "
Algebra.Com's Answer #293158 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"f%28x%29+=+log%286%2C+%285x+%2B+x%5E%28-5%29%29%29\"
\n" ); document.write( "First let's find the y-coordinate of \"the point indicated\":
\n" ); document.write( "\"f%281%29+=+log%286%2C+%285%281%29+%2B+%281%29%5E%28-5%29%29%29\"
\n" ); document.write( "\"f%281%29+=+log%286%2C+%285+%2B+1%29%29\"
\n" ); document.write( "\"f%281%29+=+log%286%2C+%286%29%29\"
\n" ); document.write( "f(1) = 1
\n" ); document.write( "So \"the point indicated\" is (1, 1)

\n" ); document.write( "Now we find the slope of the tangent at this point. For this we will need the first derivative Using the dervative of a logarithm plus the chain rule we get:
\n" ); document.write( "f'(x) = \"%281%2Fln%286%29%29%281%2F%285x+%2B+x%5E%28-5%29%29%29%285+%2B+%28-5%29x%5E%28-6%29%29\"
\n" ); document.write( "To find the slope at (1, 1) we need
\n" ); document.write( "f'(1) =
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( "f'(1) = \"%281%2Fln%286%29%29%281%2F%285+%2B+1%29%29%285+%2B+%28-5%29%29\"
\n" ); document.write( "f'(1) = \"%281%2Fln%286%29%29%281%2F6%29%280%29\"
\n" ); document.write( "f'(1) = 0

\n" ); document.write( "A line, with a slope of 1, though the point (1, 1) would be y = 1. This is the equation for the tangent line.

\n" ); document.write( "Here's a graph of f(x):
\n" ); document.write( "\"graph%28400%2C+400%2C+0.1%2C+3%2C+-5%2C+10%2C+ln%285x+%2B+x%5E%28-5%29%29%2Fln%286%29%29\" \n" ); document.write( "
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