SOLUTION: not sure if you can help... I need to find dy/dx at the point (0,1), if y^5=(x+2)^4 + e^x Iny-15

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Question 971378: not sure if you can help...
I need to find dy/dx at the point (0,1), if y^5=(x+2)^4 + e^x Iny-15
Answer by amarjeeth123(570) About Me  (Show Source):
You can put this solution on YOUR website!
y^5=(x+2)^4 + e^x Iny-15
Differentiating on both sides with respect to x we get,
5y^4(dy/dx)=4(x+2)^3+e^x.lny+e^x/y*(dy/dx)
Substituting x=0 and y=1 in the above we get,
5(dy/dx)=4(2)^3+e^0.ln(1)+e^0/1*(dy/dx)
5(dy/dx)=32+1(0)+(dy/dx)
4(dy/dx)=32
(dy/dx)=8
Answer=8