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I'm reading up on the derivative of $\arctan(x)$, and I understand all parts of the derivation except for the geometry section on the bottom of page 2: https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-b-implicit-differentiation-and-inverse-functions/session-15-implicit-differentiation-and-inverse-functions/MIT18_01SCF10_Ses15b.pdf

I understand we have $\tan(y) = x$. This means the ratio of the opposite side to adjacent side must be $x$, and the author chose to use the values $x$ and $1$. But why can't we use $x^2$ and $x$? It preserves the ratio ($x^2/x=x$), but it doesn't work.

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Check your calculation using $x^2$ and $x$: it should work. In this case hypotenuse is $\sqrt{x^4+x^2}=x\sqrt{x^2+1}$, and you can see this is simply the $1,x,\sqrt{x^2+1}$ triangle scaled by a factor of $x$, leading to the same $\cos$-value.

An algebraic derivation (which I like better) is as follows: since $\tan y=x$, $x^2+1=\tan^2 y+1=\frac{\cos^2 y+\sin^2 y}{\cos^2 y}=\frac1{\cos^2 y}$ so $\cos^2 y=\frac1{x^2+1}$. It's useful to remember $\tan^2t+1=\sec^2t$ where $\sec t=\frac1{\cos t}$.

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    $\begingroup$ Thank you very much! $\endgroup$
    – beginner
    Apr 7 '21 at 6:36
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If we use $x^2$ as the opposite side and $x$ as the adjacent side, the approach still works. In this case, the hypotenues is scaled by $x$ and becomes $x\sqrt{1+x^2}$, $\cos y = \frac{x}{x\sqrt{1+x^2}}=\frac{1}{\sqrt{1+x^2}}$.

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  • $\begingroup$ Thank you very much! $\endgroup$
    – beginner
    Apr 7 '21 at 6:36

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