document.write( "Question 1010141: Using Limits to Calculate a Derivative
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Algebra.Com's Answer #625609 by jim_thompson5910(35256)\"\" \"About 
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\"f%28x%29+=+1%2F%28x-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "Use f(x) to find f(x+h)\r
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\n" ); document.write( "\n" ); document.write( "\"f%28x%29+=+1%2F%28x-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"f%28x%2Bh%29+=+1%2F%28x%2Bh-1%29\"\r
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\n" ); document.write( "\n" ); document.write( "Compute the difference quotient\r
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\n" ); document.write( "\n" ); document.write( "\"%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%281%2F%28x%2Bh-1%29-1%2F%28x-1%29%29%2Fh\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%28-h%29%2F%28h%28x-1%29%28x%2Bh-1%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%28-1%29%2F%28%28x-1%29%28x%2Bh-1%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "The difference quotient is \"%28f%28x%2Bh%29-f%28x%29%29%2Fh+=+%28-1%29%2F%28%28x-1%29%28x%2Bh-1%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "Now evaluate the limit as h goes to 0. Since we no longer have to worry about division by zero errors, this is the equivalent to plugging in h = 0 to get \r
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\n" ); document.write( "\n" ); document.write( "So the derivative of f(x) is f ' (x) = \"+%28-1%29%2F%28%28x-1%29%5E2%29\"
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