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Derivatives Of Trigonometric Functions

Q: Do I need to memorize all six derivatives?

It's highly recommended to memorize the derivatives of $ \sin(x) $ and $ \cos(x) $. You can then derive the other four quickly using the quotient rule. However, for speed and efficiency on tests, memorizing all six is a very good idea.

Q: What's the difference between differentiating sin²(x) and sin(x²)?

This is a great question about the chain rule. For $ y = \sin^2(x) $, which is $ (\sin(x))^2 $, the outer function is $ u^2 $ and the inner is $ \sin(x) $, so the derivative is $ 2\sin(x)\cos(x) $. For $ y = \sin(x^2) $, the outer function is $ \sin(u) $ and the inner is $ x^2 $, so the derivative is $ \cos(x^2) \cdot 2x $.

Q: How do you find the derivative of an inverse trig function like arcsin(x)?

Derivatives of inverse trigonometric functions like $ \arcsin(x) $ or $ \arctan(x) $ are a separate but related topic in calculus. They are found using a technique called implicit differentiation and result in algebraic expressions. For example, the derivative of $ \arcsin(x) $ is $ \frac{1}{\sqrt{1-x^2}} $.

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Welcome to the exciting world where calculus meets trigonometry! This lesson will guide you through finding the derivatives of trigonometric functions, a fundamental skill for understanding rates of change in cyclical phenomena like waves, oscillations, and circular motion. Let's dive in!

Derivatives Of Trigonometric Functions — an original Algebra911 reference diagram defining derivatives of trigonometric functions with its key formula and a worked example. — Derivatives Of Trigonometric Functions

What Are Trigonometric Derivatives?

The derivatives of trigonometric functions are the rules and formulas we use to find the instantaneous rate of change of functions like sine, cosine, and tangent. In simpler terms, if you graph a function like a sine wave, its derivative tells you the exact slope of the curve at any given point. This is a crucial concept in calculus because trigonometric functions model countless real-world phenomena, from the vibration of a guitar string to the alternating current in our homes.

You might recall that the derivative of a function gives us the slope of the tangent line. For the wavy, oscillating graphs of trig functions, this slope is constantly changing. At a peak of the sine curve, the slope is zero. As the curve goes downwards, the slope becomes negative. The derivative is a new function that precisely describes this changing slope. The six fundamental trigonometric functions—sine, cosine, tangent, cosecant, secant, and cotangent—each have their own unique derivative formula.

How Do We Find the Derivative of Sine?

Instead of just memorizing a formula, let's understand where it comes from. We can find the derivative of $f (x) = \sin (x)$ using the limit definition of a derivative:

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

Let's substitute $\sin (x)$ into this definition:

$\frac{d}{d x} \sin (x) = lim_{h \to 0} \frac{\sin (x + h) - \sin (x)}{h}$

To solve this, we need the sine angle addition formula: $\sin (A + B) = \sin (A) \cos (B) + \cos (A) \sin (B)$ . Applying this to $\sin (x + h)$ :

$= lim_{h \to 0} \frac{(\sin (x) \cos (h) + \cos (x) \sin (h)) - \sin (x)}{h}$

Now, we can rearrange the terms in the numerator to group the $\sin (x)$ terms together:

$= lim_{h \to 0} \frac{\sin (x) \cos (h) - \sin (x) + \cos (x) \sin (h)}{h}$

Let's factor out $\sin (x)$ from the first two terms and split the limit into two parts:

$= lim_{h \to 0} \frac{\sin (x) (\cos (h) - 1) + \cos (x) \sin (h)}{h}$

$= lim_{h \to 0} (\frac{\sin (x) (\cos (h) - 1)}{h} + \frac{\cos (x) \sin (h)}{h})$

$= lim_{h \to 0} (\sin (x) \cdot \frac{\cos (h) - 1}{h}) + lim_{h \to 0} (\cos (x) \cdot \frac{\sin (h)}{h})$

At this point, we rely on two fundamental trigonometric limits (which are proven in more advanced texts and require angles to be in radians):

$lim_{h \to 0} \frac{\sin (h)}{h} = 1$
$lim_{h \to 0} \frac{\cos (h) - 1}{h} = 0$

Substituting these values back into our expression:

$= (\sin (x) \cdot 0) + (\cos (x) \cdot 1)$

$= 0 + \cos (x) = \cos (x)$

And there we have it! We have formally proven that the derivative of the sine function is the cosine function.

\frac{d}{dx} \sin(x) = \cos(x)

What About the Derivative of Cosine?

We can follow a very similar process for $f (x) = \cos (x)$ . We start with the limit definition again:

$\frac{d}{d x} \cos (x) = lim_{h \to 0} \frac{\cos (x + h) - \cos (x)}{h}$

This time, we use the cosine angle addition formula: $\cos (A + B) = \cos (A) \cos (B) - \sin (A) \sin (B)$ .

$= lim_{h \to 0} \frac{(\cos (x) \cos (h) - \sin (x) \sin (h)) - \cos (x)}{h}$

Again, we group the $\cos (x)$ terms and split the limit:

$= lim_{h \to 0} \frac{\cos (x) (\cos (h) - 1) - \sin (x) \sin (h)}{h}$

$= lim_{h \to 0} (\cos (x) \cdot \frac{\cos (h) - 1}{h}) - lim_{h \to 0} (\sin (x) \cdot \frac{\sin (h)}{h})$

Using the same two special limits as before:

$= (\cos (x) \cdot 0) - (\sin (x) \cdot 1)$

$= 0 - \sin (x) = - \sin (x)$

This gives us our second fundamental trigonometric derivative. Notice the crucial negative sign!

\frac{d}{dx} \cos(x) = -\sin(x)

How Are the Other Four Trig Derivatives Found?

Fortunately, we don't need to go through the limit definition for the other four functions. We can find their derivatives by expressing them in terms of sine and cosine and then applying the Quotient Rule, which states:

\frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}

Let's find the derivative of $\tan (x)$ . First, we rewrite it as $\tan (x) = \frac{\sin (x)}{\cos (x)}$ . Here, $f (x) = \sin (x)$ and $g (x) = \cos (x)$ . We already know their derivatives: $f^{'} (x) = \cos (x)$ and $g^{'} (x) = - \sin (x)$ . Now, apply the quotient rule:

$\frac{d}{d x} \tan (x) = \frac{(\cos (x)) (\cos (x)) - (\sin (x)) (- \sin (x))}{(\cos (x))^{2}}$

$= \frac{\cos^{2} (x) + \sin^{2} (x)}{\cos^{2} (x)}$

Using the Pythagorean identity $\sin^{2} (x) + \cos^{2} (x) = 1$ , the numerator simplifies:

$= \frac{1}{\cos^{2} (x)}$

Since $\sec (x) = \frac{1}{\cos (x)}$ , we can write the final result as:

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

We can use the same technique for the remaining three functions: $\cot (x) = \frac{\cos (x)}{\sin (x)}$ , $\sec (x) = \frac{1}{\cos (x)}$ , and $\csc (x) = \frac{1}{\sin (x)}$ . Try deriving them yourself as practice! The results are summarized in a later section.

Putting It All Together: Worked Examples

The real power of these rules comes when we combine them with other differentiation rules like the Product, Quotient, and Chain Rules.

Example 1

Find the derivative of $f (x) = x^{3} \cos (x)$ .

Solution: This is a product of two functions: $x^{3}$ and $\cos (x)$ . We must use the Product Rule: $(u v)^{'} = u^{'} v + u v^{'}$ .

Let $u = x^{3}$ , so $u^{'} = 3 x^{2}$ .
Let $v = \cos (x)$ , so $v^{'} = - \sin (x)$ .

Now, we plug these into the formula:

$f^{'} (x) = (3 x^{2}) (\cos (x)) + (x^{3}) (- \sin (x))$

$f^{'} (x) = 3 x^{2} \cos (x) - x^{3} \sin (x)$

Example 2

Find the derivative of $g (t) = \frac{\sin (t)}{t^{2} + 1}$ .

Solution: This is a quotient, so we use the Quotient Rule: $(\frac{f}{g})^{'} = \frac{f^{'} g - f g^{'}}{g^{2}}$ .

Let the numerator be $f (t) = \sin (t)$ , so $f^{'} (t) = \cos (t)$ .
Let the denominator be $g (t) = t^{2} + 1$ , so $g^{'} (t) = 2 t$ .

Substitute into the formula:

$g^{'} (t) = \frac{(\cos (t)) (t^{2} + 1) - (\sin (t)) (2 t)}{(t^{2} + 1)^{2}}$

Simplifying the numerator gives the final answer:

$g^{'} (t) = \frac{(t^{2} + 1) \cos (t) - 2 t \sin (t)}{(t^{2} + 1)^{2}}$

Example 3

Find the derivative of $h (x) = \tan (4 x^{2} - 5)$ .

Solution: This requires the Chain Rule. We have an "outside" function, $\tan (u)$ , and an "inside" function, $u = 4 x^{2} - 5$ .

The Chain Rule states $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ .

Differentiate the outside function: The derivative of $\tan (u)$ is $\sec^{2} (u)$ . So we get $\sec^{2} (4 x^{2} - 5)$ .
Differentiate the inside function: The derivative of $4 x^{2} - 5$ is $8 x$ .
Multiply them together:

$h^{'} (x) = \sec^{2} (4 x^{2} - 5) \cdot (8 x)$

It's conventional to write the polynomial part first:

$h^{'} (x) = 8 x \sec^{2} (4 x^{2} - 5)$

Key formulas for derivatives of trigonometric functions by Algebra911.

Quick Reference: Table of Trigonometric Derivatives

For quick review and memorization, here are the six trigonometric derivatives. Notice the patterns: the derivatives of the "co-" functions (cosine, cotangent, cosecant) are all negative.

Function $f (x)$	Derivative $f^{'} (x)$
$\sin (x)$	$\cos (x)$
$\cos (x)$	$- \sin (x)$
$\tan (x)$	$\sec^{2} (x)$
$\csc (x)$	$- \csc (x) \cot (x)$
$\sec (x)$	$\sec (x) \tan (x)$
$\cot (x)$	$- \csc^{2} (x)$

What Are Some Common Mistakes to Avoid?

When learning these new rules, it's easy to make a few common errors. Being aware of them is the first step to avoiding them!

The Sign Error: The most frequent mistake is forgetting the negative sign for the derivatives of cosine, cotangent, and cosecant. A good mnemonic is that if the function starts with "co-", its derivative is negative.
Forgetting the Chain Rule: This is a huge one. The derivative of $\sin (5 x)$ is not $\cos (5 x)$ . You must apply the chain rule: differentiate the outside (sine becomes cosine) and multiply by the derivative of the inside (the derivative of $5 x$ is $5$ ). The correct answer is $5 \cos (5 x)$ .
Mixing Up Formulas: The derivatives of secant and cosecant can be tricky to remember. Students often mix them up or forget which one is negative. Similarly, it's easy to confuse the derivative of tangent ( $\sec^{2} (x)$ ) with the derivative of secant ( $\sec (x) \tan (x)$ ). Practice and using the summary table are key.
Incorrectly Applying the Quotient Rule: When deriving tangent or secant, a small mistake in the quotient rule (like getting the order of subtraction wrong in the numerator) will lead to an incorrect result. Write out each part of the rule carefully.
Using Degrees Instead of Radians: All the derivative formulas we've discussed are only valid when $x$ is measured in radians. This is because the fundamental limits used in the proofs (like $lim_{h \to 0} \frac{\sin (h)}{h} = 1$ ) only hold true for radians.

Frequently Asked Questions

Why is the derivative of sin(x) equal to cos(x)?

The formal reason comes from the limit definition of a derivative, which shows that the rate of change of the sine function at any point x is exactly equal to the value of the cosine function at that same point. Geometrically, if you look at the sine wave, its steepest positive slope occurs at x=0, where cos(0)=1, and its slope is zero at its peak (x=π/2), where cos(π/2)=0.

Do the angles have to be in radians?

Yes, absolutely. All the standard calculus formulas for trigonometric derivatives require the angle variable to be in radians. This is because the proofs rely on fundamental limits that are only true when using radian measure. Using degrees will give an incorrect answer.

Do I need to memorize all six derivatives?

It's highly recommended to memorize the derivatives of $\sin (x)$ and $\cos (x)$ . You can then derive the other four quickly using the quotient rule. However, for speed and efficiency on tests, memorizing all six is a very good idea.

What's the difference between differentiating sin²(x) and sin(x²)?

This is a great question about the chain rule. For $y = \sin^{2} (x)$ , which is $(\sin (x))^{2}$ , the outer function is $u^{2}$ and the inner is $\sin (x)$ , so the derivative is $2 \sin (x) \cos (x)$ . For $y = \sin (x^{2})$ , the outer function is $\sin (u)$ and the inner is $x^{2}$ , so the derivative is $\cos (x^{2}) \cdot 2 x$ .

What are these derivatives used for in the real world?

Trigonometric derivatives are essential in physics and engineering. They are used to analyze any form of oscillation or wave motion, such as simple harmonic motion (springs, pendulums), alternating current (AC) circuits, sound waves, and light waves. They help us calculate velocity and acceleration in these systems.

Is there a pattern for the derivatives of 'co-' functions?

Yes, there is a helpful pattern. The derivative of each 'co-' function (cosine, cotangent, cosecant) is negative. This can help you remember to include the minus sign when differentiating them.

How do you find the derivative of an inverse trig function like arcsin(x)?

Derivatives of inverse trigonometric functions like $\arcsin (x)$ or $\arctan (x)$ are a separate but related topic in calculus. They are found using a technique called implicit differentiation and result in algebraic expressions. For example, the derivative of $\arcsin (x)$ is $\frac{1}{\sqrt{1 - x^{2}}}$ .