Question

How do you find `dy/dx` by implicit differentiation of `x=sec(1/y)`?

Answer 1

Given:

`x=sec(1/y)`

Differentiate both sides:

`(d(x))/dx=(d(sec(1/y)))/dx`

The left side is 1:

`1=(d(sec(1/y)))/dx`

The right side requires the recursive use of the chain rule:

let `u = 1/y` then

`(d(sec(1/y)))/dx = (dsec(u))/(du) (du)/dydy/dx`

`(dsec(u))/(du) = tan(u)sec(u)`

Reverse the substitution:

`(dsec(u))/(du) = tan(1/y)sec(1/y)`

`(du)/dy = -1/y^2`

Returning to the equation:

`1=tan(1/y)sec(1/y)(-1)/y^2dy/dx`

Solving for `dy/dx` in 3 steps:

`-y^2=tan(1/y)sec(1/y)dy/dx`

`-y^2cot(1/y)= sec(1/y)dy/dx`

`dy/dx = -y^2cot(1/y)cos(1/y)`