How do I find dy/dx if (1) y = square root of (k^2 + u^2) and if (2) x = square root of (k^2 + y^2)?
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Answer:
Use chain rule, i) y' = [1/ 2sqrt(k^2 + tan^2 x)] [2tan x] [sec^2 x] = tan x sec^2 x/sqrt(k^2 + tan^2 x) ii) 1 = [1/ 2sqrt(k^2 + y^2)][2yy'] y' = sqrt(k^2 + y^2)/y
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Other answers
Use chain rule, i) y' = [1/ 2sqrt(k^2 + tan^2 x)] [2tan x] [sec^2 x] = tan x sec^2 x/sqrt(k^2 + tan^2 x) ii) 1 = [1/ 2sqrt(k^2 + y^2)][2yy'] y' = sqrt(k^2 + y^2)/y
sahsjing
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