# Understanding a quote about implicit differentiation

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Summary:
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:
Inverse tangent function derivative can be also obtained by implicit differentiation: if y=tan−1⁡x, then x=tan⁡y, and 1=(sec2⁡y)dydx=(1+tan2⁡y)dydx=(1+x2)dydx Hence, ddxtan−1⁡x=11+x2
It's about advice on Lebniz's notation1=(sec2⁡y)dydx means dxdx=(sec2⁡y)dydx, I'm quite sure. Why (sec2⁡y)dydx=(1+tan2⁡y)dydx? But I'm also quite sure that the right notation for (sec2⁡y)dydx=(1+tan2⁡y)dydx would be (sec2⁡y)ddx=(1+tan2⁡y)ddx

PeroK
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##sec^2y = 1 + tan^2 y## is one of the most important trig identities.

• mcastillo356
Gold Member
Sorry, I've posted nonsense: I was trying to be clever about Leibniz's notation: impose my personal point of view. I've got troubles posting and editing. My weird opinion was... Well, I will post again, and then my unfounded opinion:

Quote from the book:

The derivative of the inverse tangent function can be calculated also by implicit differentiation: if ##y=\tan^{-1} x##, then ##x=\tan y##, and

$$1=(\sec^2 y)\dfrac{dy}{dx}=(1+\tan^2 y)\dfrac{dy}{dx}=(1+x^2)\dfrac{dy}{dx}$$

Hence

$$\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$$

My botched job: set notations like ##(\sec^2 y)\dfrac{d}{dx}##

I am not native. Forgive my English.

Greetings!

• jedishrfu
Gold Member
Sorry, I've posted nonsense: I was trying to be clever about Leibniz's notation: impose my personal point of view. I've got troubles posting and editing. My weird opinion was... Well, I will post again, and then my unfounded opinion:

Quote from the book:

The derivative of the inverse tangent function can be calculated also by implicit differentiation: if ##y=\tan^{-1} x##, then ##x=\tan y##, and

$$1=(\sec^2 y)\dfrac{dy}{dx}=(1+\tan^2 y)\dfrac{dy}{dx}=(1+x^2)\dfrac{dy}{dx}$$

Hence

$$\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$$

My botched job: set notations like ##(\sec^2 y)\dfrac{d}{dx}##

I am not native. Forgive my English.

Greetings!

• jedishrfu and Delta2
PeroK
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