document.write( "Question 1161974: Let $\"y+=%28f%28u%29+%2B+4x%29%5E2\"$ and $\"u+=+x%5E3+-+2x\"$ . If $\"f%284%29+=+7\"$ and $\"dy%2Fdx+=+16\"$ when $\"x+=+2\"$ , find f '(4). \n" ); document.write( "

Algebra.Com's Answer #785665 by jim_thompson5910(35256) $\"\"$ $\"About$

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\n" ); document.write( "When x = 2,
\n" ); document.write( "u = x^3 - 2x
\n" ); document.write( "u = (2)^3 - 2(2)
\n" ); document.write( "u = 4
\n" ); document.write( "f(u) = f(4) = 7
\n" ); document.write( "So x = 2 leads to f(u) = 7\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "y = ( f(u) + 4x )^2
\n" ); document.write( "dy/dx = 2( f(u) + 4x )*( f'(u) + 4 ) ... chain rule
\n" ); document.write( "16 = 2( 7 + 4(2) )*( f'(u) + 4 ) ... substitution; isolate f'(u)
\n" ); document.write( "16 = 2( 15 )*( f'(u) + 4 )
\n" ); document.write( "16 = 30( f'(u) + 4 )
\n" ); document.write( "16 = 30*f'(u) + 120
\n" ); document.write( "16-120 = 30*f'(u) ... subtract 120 from both sides
\n" ); document.write( "-104 = 30*f'(u)
\n" ); document.write( "30*f'(u) = -104
\n" ); document.write( "f'(u) = -104/30 ... divide both sides by 30
\n" ); document.write( "f'(u) = -52/15 ... reduce
\n" ); document.write( "f'(4) = -52/15 ... recall that x = 2 leads to u = 4\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Answer: -52/15
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