SOLUTION: Please help me solve this equation: 1. Determine dy/dx for y = x√x(5x^2 − 3)^4 (There is no need to simplify the answer.) 2. For the curve defined by the equation

Algebra ->  Equations -> SOLUTION: Please help me solve this equation: 1. Determine dy/dx for y = x√x(5x^2 − 3)^4 (There is no need to simplify the answer.) 2. For the curve defined by the equation      Log On


   

Question 1078501: Please help me solve this equation:
1. Determine dy/dx for y = x√x(5x^2 − 3)^4
(There is no need to simplify the answer.)
2. For the curve defined by the equation x^2 − xy + y^2 = 3
(a) Show that
dy/dx =2x − y/x − 2y
(b) Find the equation of the normal to the curve at the point (−1, 1).
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
y=f%2Ag where f=x and g=sqrt%28x%285x%5E2-3%29%5E4%29
So then,
dy%2Fdx=f%2A%28dg%2Fdx%29%2Bg%2A%28df%2Fdx%29
df%2Fdx=1






So then,

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2a. Differentiate implicitly,
2xdx-%28xdy%2Bydx%29%2B2ydy=0
%28-x%2B2y%29dy%2B%282x-y%29dx=0
%28x-2y%29dy=%282x-y%29dx
dy%2Fdx=%282x-y%29%2F%28x-2y%29
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2b. The slope of the tangent line is equal to the value of the derivative at the point.
m%5Bt%5D=+%282%28-1%29-1%29%2F%28-1-2%281%29%29+=+%28-3%29%2F%28-3%29=1
The tangent and normal line are perpendicular to each other so their slopes are negative reciprocals.
m%5Bn%5D%2Am%5Bt%5D=-1
m%5Bn%5D=-1
So then using the point-slope form of a line,
y-1=-1%28x-%28-1%29%29
y-1=-%28x%2B1%29
y-1=-x-1
y%5Bn%5D=-x
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