You are watching: Why is the derivative of tanx sec2x

This question does not meet nlinux.orgematics Stack Exreadjust guidelines. It is not currently accepting answers.

**Want to improve this question?**Upday the question so it's on-topic for nlinux.orgematics Stack Exchange.

Closed 2 years ago.

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

Cite

Follow

answered Jul 24 "19 at 9:16

Hans LundmarkHans Lundmark

44.9k77 gold badges7272 silver badges132132 bronze badges

$endgroup$

Add a comment |

3

$egingroup$

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.

See more: Nick Reineke On Twitter: " Grounded In Reality And Full Of Real World Problems

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.