document.write( "Question 1078501: Please help me solve this equation:\r
\n" ); document.write( "\n" ); document.write( "1. Determine dy/dx for y = x√x(5x^2 − 3)^4
\n" ); document.write( "(There is no need to simplify the answer.)
\n" ); document.write( "2. For the curve defined by the equation x^2 − xy + y^2 = 3
\n" ); document.write( "(a) Show that
\n" ); document.write( "dy/dx =2x − y/x − 2y
\n" ); document.write( "(b) Find the equation of the normal to the curve at the point (−1, 1).
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Algebra.Com's Answer #692947 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
\"y=f%2Ag\" where \"f=x\" and \"g=sqrt%28x%285x%5E2-3%29%5E4%29\"
\n" ); document.write( "So then,
\n" ); document.write( "\"dy%2Fdx=f%2A%28dg%2Fdx%29%2Bg%2A%28df%2Fdx%29\"
\n" ); document.write( "\"df%2Fdx=1\"
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\n" ); document.write( "So then,
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\n" ); document.write( "2a. Differentiate implicitly,
\n" ); document.write( "\"2xdx-%28xdy%2Bydx%29%2B2ydy=0\"
\n" ); document.write( "\"%28-x%2B2y%29dy%2B%282x-y%29dx=0\"
\n" ); document.write( "\"%28x-2y%29dy=%282x-y%29dx\"
\n" ); document.write( "\"dy%2Fdx=%282x-y%29%2F%28x-2y%29\"
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\n" ); document.write( "2b. The slope of the tangent line is equal to the value of the derivative at the point.
\n" ); document.write( "\"m%5Bt%5D=+%282%28-1%29-1%29%2F%28-1-2%281%29%29+=+%28-3%29%2F%28-3%29=1\"
\n" ); document.write( "The tangent and normal line are perpendicular to each other so their slopes are negative reciprocals.
\n" ); document.write( "\"m%5Bn%5D%2Am%5Bt%5D=-1\"
\n" ); document.write( "\"m%5Bn%5D=-1\"
\n" ); document.write( "So then using the point-slope form of a line,
\n" ); document.write( "\"y-1=-1%28x-%28-1%29%29\"
\n" ); document.write( "\"y-1=-%28x%2B1%29\"
\n" ); document.write( "\"y-1=-x-1\"
\n" ); document.write( "\"y%5Bn%5D=-x\"
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