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I understand also why the derivative of \$sin x\$ is \$cos x\$, and why the derivative of \$cos x\$ is \$-sin x\$ in a geometric means.

But I can not understand why the derivative of \$ an x\$ is \$sec^2 x\$.

Can someone please describe this in a visual, geometric means making use of the unit circle?

The complying with is maybe a small little hand-wavy, however the idea is hopefully clear.

If \$ heta\$ is the angle \$EAH\$ in the number, then \$ an heta\$ is the lenght of the segment \$BC\$. If we boost that angle slightly (or “infinitesimally”), then the segment \$EF\$ is around \$d heta\$, and also we would prefer to understand the size of \$CD\$, given that that"s the change in \$ an heta\$.

Since the triangle \$AHE\$ is \$cos heta\$ times smaller than the triangle \$ABC\$, the segment \$EF\$ is \$cos heta\$ times shorter than \$CG\$. And \$CG\$ is consequently \$cos heta\$ times shorter than \$CD\$ (similar triangles; the angle \$GCD\$ is \$ heta\$ and the angle \$CGD\$ is \$pi/2\$ approximately an infinitesimal correction). So \$EF\$ is \$cos^2 heta\$ times shorter than \$CD\$, and thus\$\$d( an heta ) = CD = fracEFcos^2 heta = fracd hetacos^2 heta.\$\$

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answered Jul 24 "19 at 9:16

Hans LundmarkHans Lundmark