document.write( "Question 355601: Please help me solve this equation:\r
\n" ); document.write( "\n" ); document.write( "a) Differentiate \"+f%28x%29+=+cos%5E%28-1%29+%28e%5Ex%29%2F+root%283%2C+x%29+\" DO NOT SIMPLIFY
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\n" ); document.write( "b) Simplify \"+sin+%28+tan%5E%28-1%29+%282x%29+%29+\"
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Algebra.Com's Answer #253977 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"+f%28x%29+=+cos%5E%28-1%29+%28e%5Ex%29%2F+root%283%2C+x%29+\"
\n" ); document.write( "This problem uses the several levels of the chain rule. First we will
\n" ); document.write( "Let u = \"cos%5E%28-1%29+%28e%5Ex%29\" and \"v+=+root%283%2C+x%29+=+x%5E%281%2F3%29\".
\n" ); document.write( "This makes f = u/v and from this we know that
\n" ); document.write( "f' = (v*u' - u*v')/v^2
\n" ); document.write( "For this we will need u' and v':
\n" ); document.write( "u' = \"%281%2Fsqrt%281+-+%28e%5Ex%29%5E2%29%29%2A%28d%2Fdx%29%28e%5Ex%29\"
\n" ); document.write( "Since the derivative of \"e%5Ex\" is \"e%5Ex\" this becomes:
\n" ); document.write( "u' = \"%281%2Fsqrt%281-e%5E%282x%29%29%29%2Ae%5Ex+=+e%5Ex%2Fsqrt%281-e%5E%282x%29%29\"
\n" ); document.write( "and v' = \"%281%2F3%29x%5E%281-%281%2F3%29%29+=+%281%2F3%29x%5E%28-2%2F3%29\"
\n" ); document.write( "Substituting u, u', v and v' into the f' equation we get:
\n" ); document.write( "f' =
\n" ); document.write( "And if you're not supposed to simplify, then I guess this mess is an acceptable answer.

\n" ); document.write( "b) \"+sin+%28+tan%5E%28-1%29+%282x%29+%29+\"
\n" ); document.write( "For this one, picture a right triangle. For one of the acute angles we want the tangent to be 2x. In other words we want the ratio of opposite/adjacent to be 2x. An opposite side of 2x and an adjacent side of 1 would give us this ratio. We want to find the sin of this angle. Since sin is opposite/hypotenuse, we will need an expression for the hypotenuse. For this we can use the Pythagorean Theorem:
\n" ); document.write( "Opposite^2 + Adjacent^2 = Hypotenuse^2
\n" ); document.write( "Putting our expressions for the opposite side and adjacent side into this we get:
\n" ); document.write( "\"%282x%29%5E2+%2B+%281%29%5E2\" = Hypotenuse^2
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"4x%5E2+%2B+1\" = Hypotenuse^2
\n" ); document.write( "So the Hypotenuse is \"sqrt%284x%5E2+%2B+1%29\"
\n" ); document.write( "Now we can express the sin ratio:
\n" ); document.write( "\"+sin+%28+tan%5E%28-1%29+%282x%29+%29+=+%282x%29%2Fsqrt%284x%5E2+%2B+1%29\"
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