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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.\$\$ Share
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answered Jul 24 "19 at 9:16 Hans LundmarkHans Lundmark
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First, look at the graph of \$ an(x)\$. It has actually vertical asymptotes at integer multiples of \$dfracpi2\$ and is unidentified at \$-dfracpi2\$ and also \$dfracpi2\$. Observe that from \$-dfracpi2 leq x leq dfracpi2\$, the slope of \$ an(x)\$ is constantly enhancing. Notice that it increases faster from \$-dfracpi2 leq x leq -dfracpi4\$ and also from \$dfracpi4 leq x leq dfracpi2\$. Then, let"s take a look at the graph of \$sec^2(x)\$. Observe that \$sec^2(x)\$ is constantly positive as the slope of \$ an(x)\$ was always positive. Also observe that \$sec^2(x)\$ has actually vertical asymptotes at integer multiples of \$dfracpi2\$. If we zone in on \$-dfracpi2 leq x leq dfracpi2\$, then we view that the value of \$sec^2(x)\$ is higher as we method \$x=-dfracpi2\$ or \$x=dfracpi2\$. This is because we deserve to think of the derivative as slope and also formerly observed that the slope was best near the asymptotes.

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As such, it is organic for \$sec^2(x)\$ to be the derivative of \$ an(x)\$. The very same technique will certainly work-related for \$sin(x), cos(x)\$, and many others. If you are uncomfortable via the algebra then it is finest attract a function and its derivative on graph paper.