Saturday, July 20, 2019

Calculus: Early Transcendentals, Chapter 3, 3.5, Section 3.5, Problem 20

d/(dx) tan(x-y) =d/(dx)(y/(1+x^2))
sec^2(x-y) *d/(dx)(x-y) = d/(dx) y(1+x^2)^-1
(1-dy/dx)sec^2(x-y) = y(-1)(1+x^2)^(-2)(2x) +(1+x^2)^-1 dy/dx
(1-dy/dx)sec^2(x-y) = (-2xy)/(1+x^2)^2 + (1/(1+x^2)) dy/dx
(sec^2(x-y) +(1/(1+x^2))) dy/dx = sec^2(x-y) + (2xy)/(1+x^2)^2
(((1+x^2)sec^2(x-y) + 1)/(1+x^2)) dy/dx = ((1+x^2)^2sec^2(x-y) +2xy)/(1+x^2)^2
dy/dx=(((1+x^2)^2sec^2(x-y) +2xy))/((1+x^2)((1+x^2)sec^2(x-y) +1))

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