Calculus
posted by Anonymous .
y=arctan(sqrt(9x^21))^2
dy/dx=?

since (√(9x^21))^2 = 9x^21,
I assume you mean arctan^2 √(9x^21)
let u = 9x^21
y = arctan^2 (√u)
y' = 2 arctan(√u) * 1/(u+1) * 1/(2√u) u'
= 2 arctan(√9x^21) * 1/(9x^2) * 1/(2√(9x^21)) * 18x
= 2 arctan(√9x^21) / [x √(9x^21)]
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