How would you find the derivative of sin(arctan(x)) ?
1) You have understood that “sin(arc tan x) = x/(1+x^2)^1/2 2) Now requires differentiating using u/v form [Derivative of u/v form = {(u’)(v) – (u)(v ‘)}/(v^2)] 3) Derivative of [x/(1+x^2)^1/2] = [(1+x^2)^(1/2) – (x)*(2x)/2(1+x^2)^(1/2)]/(1+x^2) 4) Multiplying both numerator and denominator, by (1+x^2)^1/2, we get, ==> [1 + x^2 – x^2]/(1+x^2)^3/2 = 1/{(1 + x^2)^(3/2)} But instead of the above, you may do in another way: Directly differentiating, derivative of “sin(arc tan x) = cos(arc tan x)*{derivative of (arc tan x}} ==> dy/dx = cos(arc tan x)/(1 + x^2) cos(arc tan x) = 1/(1 + x^2)^(1/2) ==> dy/dx = 1/(1 + x^2)^(3/2)
