Find the equation of the normal to the graph at M(2,1)

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SUMMARY

The equation of the normal to the graph defined by the implicit function \(2x^3 + 8\ln y + x^3 e^y - 16 - 8e = 0\) at the point M(2,1) is derived using implicit differentiation. The derivative \(\dot{y}\) is calculated as \(-\frac{x^2(6 + 3e^y)}{\frac{8}{y} + x^3 e^y}\). At the point (2,1), the slope of the normal line is determined to be \(m(N) = \frac{2(1+e)}{3(2+e)}\), leading to the equation \(Y - 1 = \frac{2(1+e)}{3(2+e)}(x - 2)\). The calculations and the resulting equation are confirmed to be correct.

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Dell
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if:
2x3+8lny+x3ey-16-8e=0

what i did was
dy/dx=[tex]\dot{y}[/tex]

6x2+8(1/y)[tex]\dot{y}[/tex]+3x2ey+x3ey[tex]\dot{y}[/tex]=0

[tex]\dot{y}[/tex]=-(x2(6+3ey)/(8/y+x3ey)

then to find the incline of the normal, m=-1/[tex]\dot{y}[/tex]

so at(2,1)

[tex]\dot{y}[/tex]=-3(2+e)/2(1+e)

m(N)=2(1+e)/3(2+e)

does this seem right??

Y-1=[2(1+e)/3(2+e)]{x-2}
 
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Yes, it's correct.
 

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