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2x^{3}+8lny+x^{3}e^{y}-16-8e=0

what i did was

dy/dx=[tex]\dot{y}[/tex]

6x^{2}+8(1/y)[tex]\dot{y}[/tex]+3x^{2}e^{y}+x^{3}e^{y}[tex]\dot{y}[/tex]=0

[tex]\dot{y}[/tex]=-(x^{2}(6+3e^{y})/(8/y+x^{3}e^{y})

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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# Homework Help: Find the equation of the normal to the graph at M(2,1)

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