derivative of tan 1

derivative of tan 1 I think he originally intended to do this dy dx 1 sec2y sec2y 1 tan2y tan2y x sec2y 1 x2 dy dx 1 1 x2 I seem to recall my professor forgetting how to deriving this This is what I showed him y arctanx tany x sec 2y dy dx 1 dy dx 1 sec 2y Since tany x 1 and sqrt 1 2 x 2 sqrt 1 x 2 sec

1 Answer I m assuming you are thinking of this as being a function of two independent variables x and y z tan 1 y x The answers are frac partial z partial x frac y x 2 y 2 and frac partial z partial y frac x x 2 y 2 Both of these facts can be derived with the Chain Rule the Power Rule and the Remember that you have which means that you get Finally replace tan2y with x2 to get y 2x 1 x 4 You can differentiate a function y tan 1 x 2 by using implicit differentiation So if you have a function y tan 1 x 2 then you know that you can write tan y x 2 Differentiate both sides with respect to x to get d

derivative of tan 1

derivative of tan 1
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Derivative of tan 1 f x Ask Question Asked 8 years 2 months ago Modified 8 years 2 months ago The derivative is 1 x 2 1 d dx arctan x 1 1 x 2 So d dx arctan u 1 1 u 2 du dx And d dx arctan 1 x 1 1 1 x 2 d dx 1 x 1 1 1 x 2 1

Explanation u tan 1 y x This problem needs a slight prerequisite of chain rule and quotient rule x 2 y xdy dx x 2 y 2 Thus d dx tan 1 y x x 2 y xdy dx x 2 y 2 It really depends upon what you are doing and which independent variables matter to you Let z x y tan 1 y x qquad qquad implies tan z y x qquad triangle Derivative of tan 1 left sqrt frac a b a b tan frac x2 right Hot Network Questions Need help in identifying and learning to identify this unknown protocol which has a good chance to be proprietary to the hardware I m analyzing

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Terms Help Use the derivative of tan 1 and the chain rule The derivative of tan 1x is 1 1 x 2 for why see note below So applying the chain rule we get d dx tan 1u 1 1 u 2 du dx In this question u 2x so we get d dx tan 1 2x 1 1 2x 2 d dx 2x 2 1 4x 2 Note If y tan 1x then tany x Explanation tanx 1 1 tanx cotx Now use the memorized d dx cotx csc2x or rewrite another step cotx cosx sinx and use the quotient rule to get d dx cotx sin2x cos2x sin2x 1 sin2x csc2x Answer link

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