Find the equation of the line tangent to the curve
xy^4+xy=780
at the point (3,4).
First, you want to find an expression for dy/dx in terms of x and y using the implicit differentiation:
i got (-y^4-y)/(x (4 y^3+1))
Then, the equation of the tangent line could be written as
y=?

Question

Find the equation of the line tangent to the curve 
xy^4+xy=780 
at the point (3,4).
First, you want to find an expression for dy/dx in terms of x and y using the implicit differentiation: 
i got (-y^4-y)/(x (4 y^3+1))
Then, the equation of the tangent line could be written as 
y=?

hartnn · Answer

u put the point (3,4) in dy/dx
x=3, y=4

hartnn · Answer

then dy/dx will give you slope of tangent

anonymous · Answer

(-4^4-4)/(3 (4 4^3+1))=-260/771

anonymous · Answer

(y-4)=(-260/771)(x-3)=y-4=mx+260/257

hartnn · Answer

(y-4)=(-260/771)(x-3)
is correct

anonymous · Answer

y=(-260/771)x+1288/257

anonymous · Answer

sweet thanks

hartnn · Answer

welcome ^_^