Introduction

Hopefully you’re excited, because we are at the end of Unit 2! We will be going over the rest of the trigonometric functions and applying what we’ve learned to do some, well you’ll see!

Let’s finish the unit strong and dive right in!

Derivative of Tangent

If $f(x)=\tan (x)$, what is the value of $f’(0)$?

We haven’t learnt what the derivative of $\tan (x)$ is, but we can solve this question through info we already know! 

If you still remember, $\tan (x)$ is simply $\frac{\sin (x)}{\cos (x)}$. If you haven’t forgotten what the previous article taught you, it was the quotient rule! We are knowledgeable in how to differentiate fractions! With $\sin (x)$ and $\cos (x)$, we are fully aware of what their derivatives are! Using the normal quotient rule procedures, we get the following!

$f(x)=\sin (x)$

$g(x)=\cos (x)$

$f’(x)=\cos (x)$

$g’(x)=-\sin (x)$

We can use the quotient rule as we usually do!

$\frac{f’g-g’f}{g^2}=\frac{\cos (x)\cdot \cos (x)-(-\sin (x))(\sin (x))}{{\cos }^2(x)}=\frac{{\cos }^2(x)+{\sin }^2(x)}{{\cos }^2(x)}=\frac{1}{{\cos }^2(x)}={\sec }^2(x)$

Without having to learn a new rule or formula or whatever, we have figured out by ourselves on how to differentiate $\tan (x)$!

$\frac{d}{dx}[\tan (x)]={\sec }^2(x)$

Of course in our excitement, we cannot forget that the problem asks us to find $f’(0)$ and not just $f(x)$, so let’s proceed with that.

$f’(0)={\sec }^2(0)=(1{)}^2=1$

Derivative of Secant

Calculate $\frac{d}{dx}[\sec (x)]$

Moving on to the next derivative, we have $\sec (x)$! We can do the same exact thing we did earlier, except with the fact that $\sec (x)=\frac{1}{\cos (x)}$. Note that $f(x)=1$ is valid, because it’s still a function!

$f(x)=1$

$g(x)=\cos (x)$

$f’(x)=0$

$g’(x)=-\sin (x)$

Once again, just plug all of this into the quotient rule!

$\frac{f’g-g’f}{g^2}=\frac{(0)(\cos (x)-(-\sin (x))(1)}{{\cos }^2(x)}=\frac{\sin (x)}{{\cos }^2(x)}=\frac{\sin (x)}{\cos (x)}\cdot \frac{1}{\cos (x)}=\tan (x)\sec (x)$

There we have it!

$\frac{d}{dx}[\sec (x)]=\tan (x)\sec (x)$

Practice

You may be asking yourself why we are heading into practice when technically, we still have two more functions to go over? Well, my friend, we have figured out two of these derivatives, and now it is YOUR turn to figure out the derivatives of the other two functions: $\csc (x)$ and $\cot (x)$! 

The unit test will do the job of making sure you know the derivatives of all the functions listed, and certainly future units! Have a great time in the next unit, Unit 3!