document.write( "Question 281944: if f(x) = \"ln%285x%5E2+%2B1%29%5E3\", then the derivative of f evaluated at x=1 equals:
\n" ); document.write( "a) 2 b) 4 c) 5 d) 6\r
\n" ); document.write( "\n" ); document.write( "My guess is D (6). I graphed the function. I used my table command with the change in x being .1 Then I found the value of the function at 1.9 and 2.1 Since the derivative represents the slope at a particular x-value, I found the slope around 2. The slope was 5.9595, which is close to 6.\r
\n" ); document.write( "\n" ); document.write( "I tried to take the derivative of this function, but I got lost. Here's what I tried:\r
\n" ); document.write( "\n" ); document.write( "\"ln%285x%5E2%2B1%29%5E3\"
\n" ); document.write( "=\"1%2F%285x%5E2+%2B1%29%5E3+%2A+3%285x%5E2%2B1%29%5E2%2810x%29\"
\n" ); document.write( "=\"30x%2F%285x%5E2+%2B1%29\" then the functional value at 1 is 30/6 = 5.
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Algebra.Com's Answer #204883 by jsmallt9(3759)\"\" \"About 
You can put this solution on YOUR website!
>if f(x) = \"ln%285x%5E2+%2B1%29%5E3\", then the derivative of f evaluated at x=1 equals:
\n" ); document.write( ">a) 2 b) 4 c) 5 d) 6\r
\n" ); document.write( "\n" ); document.write( ">My guess is D (6). I graphed the function. I used my table command with the >change in x being .1 Then I found the value of the function at 1.9 and 2.1\r
\n" ); document.write( "\n" ); document.write( "Why 1.9 and 2.1?\r
\n" ); document.write( "\n" ); document.write( "> Since the derivative represents the slope at a particular x-value,\r
\n" ); document.write( "\n" ); document.write( "True.\r
\n" ); document.write( "\n" ); document.write( "> I found the slope around 2.\r
\n" ); document.write( "\n" ); document.write( "Again, why a number near 2?\r
\n" ); document.write( "\n" ); document.write( "> The slope was 5.9595, which is close to 6.\r
\n" ); document.write( "\n" ); document.write( "You found the slope with the calculator? However, you're doing it there is an error. (I'll show you the actual slope at 2 later.)\r
\n" ); document.write( "\n" ); document.write( "> I tried to take the derivative of this function, but I got lost.\r
\n" ); document.write( "\n" ); document.write( "No you didn't!\r
\n" ); document.write( "\n" ); document.write( "> Here's what I tried:
\n" ); document.write( ">\"f%28x%29+=+ln%285x%5E2%2B1%29%5E3\"
\n" ); document.write( ">f'(x) = \"1%2F%285x%5E2+%2B1%29%5E3+%2A+3%285x%5E2%2B1%29%5E2%2810x%29\"
\n" ); document.write( ">f'(x) = \"30x%2F%285x%5E2+%2B1%29\"
\n" ); document.write( ">then the functional value at 1 is 30/6 = 5.\r
\n" ); document.write( "\n" ); document.write( "This is all totally correct! (Although I'd word it: \"The value of the (first) derivative at 1 is 30/6 = 5\") The answer to your problem is 5.\r
\n" ); document.write( "\n" ); document.write( "Again, I don't get your fixation on number near 2. They have nothing to do with the problem at hand. But FWIW,
\n" ); document.write( "f'(2) = which is approximately: 2.85714286
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