Differentiating Trig Functions

In this Calculus article I want to show you how to differentiate trig functions. There are a few required relationships to remember in order to do this. These all have to do with the trig identities as they are called. Read on for examples on how to solve these problems.
The relationships I mentioned earlier are:
$$ (d) sin x = cos x $$
$$ (d) cos x = -sin x $$
$$ (d) tan x = sec^2 x $$
$$ (d) csc x = csc x cot x $$
$$ (d) sec x = sec x tan x $$
$$ (d) cot x = -csc^2 x $$
Applying these rules to previous concepts like the power rule and chain rule will let us solve these trig functions. Any time we are able we will change from the inverse function name to the regular. This will make it easier to solve problems and will be good practice for doing this later on.

Problem 1

Find the derivative of:
$$ y = sin (x + b) $$
Let u = (x + b) and then y = sin u.
$$ \frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx} $$
$$ \frac{du}{dx} = 1 $$
It is 1 because the derivative of a constant is always 1.
$$ \frac{dy}{du} = cos u $$ and since $$ u = (x + b) $$ this becomes  $$ cos (x + b) $$
So now we just combine what we have done for the answer.
$$ \frac{dy}{dx} = \frac{dy}{du} * \frac{du}{dx} = cos (x + b) * 1 = cos (x + b) $$

Problem 2

Find the derivative of:
$$ y = sin ax^2 $$
When looking at this, you should immediately see that we are taking this derivative in two parts.
$$ (d) sin ax^2 and (d) ax^2 $$
$$ (d) sin ax^2 = cos ax^2 $$
$$ (d) ax^2 = 2 ax $$
Now we just combine these parts together and get our answer:
$$ 2 ax cos ax^2 $$

Problem 3

Find the derivative of:
$$ g(x) = (sin \sqrt{ln x}) 1 $$
We break this up just like we have been doing.
There are two parts just like the other problems.
$$ (sin \sqrt{ln x})  (\sqrt{ln x}) $$
We let U = $$ (\sqrt{ln x}) $$
The derivative of these two parts are:
$$ (cos \sqrt{ln x}) (\frac{1}{2x \sqrt{ln x}} $$

Problem 4

Find the derivative of:
$$ y = tan 3\theta $$
As you probably know, this must be in radians. Otherwise, treat this problem just like the others.
We will want to let $$ u = 3 \theta $$
Afterwards, that means our problem looks like:
$$ y = tan u $$
So when we put it all together the setup looks like this:
$$ \frac{dy}{d\theta} = \frac{dy}{du} * \frac{du}{d\theta} $$
We know that the first part is:
$$ \frac{dy}{d\theta} = 3 $$
Then the other part is:
$$ \frac{dy}{du} = sec^2 u $$
Now our whole answer is:
$$ sec^2 u * 3 $$
Now replace u. We get:
$$ 3 sec^2 (3\theta) $$
After these first few problems you can see we use some substitution. This is done to help keep things straight. When you get good with this you can just look at it and see the answer immediately. First though, practice with the substitution to get good with the fundamentals.

Problem 5

Find the derivative of:
$$ y = tan^2 x $$
Since our tangent has an exponent we will want to use the power rule on it. This becomes:
$$ y = tan^2 x = 2 tan x $$
Now we deal with the function itself.
$$ (d) tan x = sec^2 x $$
Then we combine our two answers into one.
$$ 2 tan x sec^2 x $$

Problem 6

Find the derivative of:
$$ y = sec^2 x $$
As before, you deal with the exponent and then the function.
I will also start using the notation [y'] to mean the derivative of an expression. This is done because its another notation and you will see it too.
So lets deal with the exponent first.
$$ y = sec^2 x = 2 sec x $$
Now its time for the function.
This is also our final answer.
$$ 2 sec x (sec x tan x) $$

Problem 7

Find the derivative of:
$$ y = cos^2  (2x) $$
Lets start with the exponent first.
$$ \frac{dy}{dx} = (2 cos 2x) \frac{d}{dx} (cos 2x) $$
Ok first part done. Now we just continue and do it one step at a time systematically. The next part is the cos function.
$$ \frac{d}{dx} cos 2x = -2 sin 2x $$
We have the (2x) part left so lets finish that one.
$$ \frac{d}{dx} 2x = 2 $$
All the small parts are done so lets combine them. We get:
$$ (2 cos 2x) (-2 sin 2x) (2) $$

Problem 8

Find the derivative of:
$$ y = tan^2 4x $$
We first work on the exponent.
$$ \frac{dy}{dx} y = tan^2 4x = 2 tan 4x $$
Now we deal with the tan function.
$$ \frac{dy}{dx} = 2 tan 4x \frac{d}{dx} tan 4x $$
$$ \frac{dy}{dx} = (2 tan 4x) (sec^2 4x) \frac{d}{dx} 4x $$
Now we do the last part.
$$ \frac{dy}{dx} = (2 tan 4x) ( sec^2 4x) (4) $$
We can make it look a littl4e nicer by getting rid of the 4.
Final answer should be:
$$ (8 tan 4x) ( sec^2 4x) $$

Problem 9

Find the derivative of:
$$ y = sec^4 3x $$
As done before, we just break this down piece by piece and start with the exponent.
$$ \frac{dy}{dx} = 4 sec^3 3x \frac{d}{dx} sec 3x $$
The sec function is next.
$$ (4 sec^3 3x) (sec 3x tan 3x) \frac{d}{dx} 3x $$
We now do the 3x part.
$$ (4 sec^3 3x)(sec 3x tan 3x)(3) $$
You can leave it like that or make it prettier like we did in the last problem. It could be simplified to:
$$ (12 sec^4 3x) (tan 3x) $$

Problem 10

Find the derivative of:
$$ y = cot^4 2x $$
Start with the exponent portion.
$$ \frac{dy}{dx} y = cot^4 2x = 4 cot^3 2x \frac{d}{dx} cot 2x $$
Now lets work on the cot function.
$$ \frac{dy}{dx} = (4 cot^3 2x) (-csc^2 2x) \frac{d}{dx} 2x $$
It's now time for the last part of this expression.
$$ \frac{dy}{dx} = (4 cot^3 2x) (-csc^2 2x) (2) $$
Now combine terms.
$$ (-8 cot^3 2x) (csc^2 2x) $$

Problem 11

Find the derivative of:
$$ y = 2 sin^3 (2x^4) $$
Start with the exponent.
$$ \frac{dy}{dx} = (6 sin^2 (2x^4))  (\frac{d}{dx}) (sin (2x^4)) $$
The sin function is next to work on.
$$ (\frac{dy}{dx}) = (6 sin^2 (2x^4))  (cos (2x^4))  \frac{d}{dx} (2x^4) $$
We do this exponent just like the others.
$$ \frac{dy}{dx} = (6 sin^2 (2x^4))  (cos (2x^4)) (8x^3) $$
After combining terms we get:
$$ (48 x^3 sin^2 (2x^4))  (cos (2x^4)) $$
That answer may seem a little crazy and it sure is. However, notice how we break up the problem piece by piece and use notation to be organized. I use the notation in this way to keep from getting confused with all the little terms and where they belong.

Problem 12

Find the derivative of:
$$ y = \frac{1}{3} sin 2x - tan^2 3x $$
We don't have an exponent so we go straight to the sine function.
$$ \frac{dy}{dx} = (\frac{1}{3} cos 2x)  - tan^2 3x $$
Keep the expressions separate on either side of the minus sign and take care of the exponent next with the tangent function.
$$ \frac{dy}{dx} = (\frac{1}{3} (2) cos 2x) - (2 tan 3x) \frac{d}{dx} tan 3x $$
Now do the tangent function.
$$ \frac{dy}{dx} = (\frac{1}{3} (2) cos 2x)  -  (2 tan 3x) (sec^2 3x)  \frac{d}{dx} 3x $$
The last part of this problem is the 3x portion and that is what we do next.
$$ \frac{dy}{dx} = (\frac{1}{3} (2) cos^2 2x)  -  (2 tan 3x)  (sec^2 3x)  (3)  $$
Combining terms gives us:
$$ (\frac{2}{3}  cos 2x)  -  (6 tan 3x)  (sec^2 3x) $$


Differentiating trig functions is not bad at all. If you are new to them, the key is to remain organized and take them step by step. Learning the identities and relationships between them is important but will come naturally with practice doing problems.