How do you find #(dy)/(dx)# given #xy=tanxy#?

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Answer 1 · 2016-10-06T10:05:03

#-y/x#

#xy=tanxy->y = y(x)# so

#f(y(x),x) = y(x)x-tan(y(x)x)=0#

now

#d/(dx)f(y(x),x) = y(x) + x y(x) - Sec(x y(x))^2 (y(x) + x y'(x)) = 0#

Solving for #y'(x)# we obtain

#dy/dx = -y/x#

We can also obtain the same result straightforward

#dy/dx = - (f_x)/(f_y) = -(y - y Sec(x y)^2)/(x - x Sec(x y)^2) = -y/x#