What is the slope of the tangent line of #e^x-1/(x+y)^2= C #, where C is an arbitrary constant, at #(3,1)#?

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Answer 1 · 2018-05-06T05:04:28

#-32e^3-1#

#d/dx(e^x-1/(x+y)^2)=d/dx(C)#

#e^x+2/(x+y)^3*d/dx(x+y)=0#

#e^x+2/(x+y)^3*(1+(dy)/dx)=0#

isolate #(dy)/dx#:

#2/(x+y)^3*(1+(dy)/dx)=-e^x#

#(1+(dy)/dx)=-e^x*(x+y)^3/2#

#(dy)/dx=-e^x*(x+y)^3/2-1#

plug in point #(3,1)# to find slope of tangent line:

#(dy)/dx# at #(3,1)=-e^3*(3+1)^3/2-1#

#(dy)/dx# at #(3,1)=-32e^3-1#