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mcastillo356

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- I've got a solved calculation of the inverse tangent function by implicit differentiation I'm trying to understand

Hi PF

A personal translation of a quote from Spanish "Calculus", by Robert A. Adams:

A personal translation of a quote from Spanish "Calculus", by Robert A. Adams:

It's about advice on Lebniz's notation1=(sec2y)dydx means dxdx=(sec2y)dydx, I'm quite sure. Why (sec2y)dydx=(1+tan2y)dydx? But I'm also quite sure that the right notation for (sec2y)dydx=(1+tan2y)dydx would be (sec2y)ddx=(1+tan2y)ddxInverse tangent function derivative can be also obtained by implicit differentiation: if y=tan−1x, then x=tany, and 1=(sec2y)dydx=(1+tan2y)dydx=(1+x2)dydx Hence, ddxtan−1x=11+x2