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

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Douglas K.
Dec 10, 2017

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)#

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