Derivatives of Basic Trigonometric Functions

The Derivatives of Sine and Cosine

Let's begin by building some intuition. What might the derivative of $f(x)=\sin (x)$ look like? Recall that the derivative's value at any point is the slope of the original function's tangent line at that point.

Look at the graph of $y=\sin (x)$.

• At $x=0$, the slope is steepest and positive. Notice that $\cos (0)=1$, its maximum value.
• At $x=\pi /2$, the sine curve has a horizontal tangent, so its slope is 0. Indeed, $\cos (\pi /2)=0$.
• At $x=\pi$, the slope is steepest and negative. Correspondingly, $\cos (\pi )=-1$, its minimum value.

This graphical evidence strongly suggests a remarkable relationship: the derivative of the sine function is the cosine function. Let's prove this formally.

The proof uses the limit definition of the derivative and relies on two fundamental trigonometric limits.

The process you just read through confirms our graphical hunch. For a more dynamic walkthrough of this proof, Professor Leonard's lecture is an excellent resource.

So, we have formally established our first two rules:

1. $\frac{d}{dx}(\sin x)=\cos x$
2. $\frac{d}{dx}(\cos x)=-\sin x$

The proof for the derivative of cosine is very similar to that of sine and is left as an exercise in your textbook (Exercise 20 in Section 3.3).

Derivatives of the Other Trigonometric Functions

Now that we have the derivatives for sine and cosine, we can find the derivatives for the other four trigonometric functions by expressing them as quotients and applying the Quotient Rule from our previous lesson.

Let's derive the derivative of the tangent function together.
We know that $\tan x=\frac{\sin x}{\cos x}$. Let's apply the Quotient Rule, ${\left(\frac{f}{g}\right)}^{\prime }=\frac{g{f}^{\prime }-f{g}^{\prime }}{g^2}$, where $f(x)=\sin x$ and $g(x)=\cos x$.

• $f^{\prime }(x)=\cos x$
• $g^{\prime }(x)=-\sin x$

Substituting these into the formula:

Using the Pythagorean identity ${\sin }^{2}x+{\cos }^{2}x=1$:

So, $\frac{d}{dx}(\tan x)={\sec }^{2}x$.

The derivatives for cosecant, secant, and cotangent can be found in the exact same way. You can see these derivations in the resources and are encouraged to try them yourself.

Here is the complete list of the six trigonometric derivatives. It is essential to memorize these.

Derivatives of Trigonometric Functions:

• $\frac{d}{dx}(\sin x)=\cos x$
• $\frac{d}{dx}(\cos x)=-\sin x$
• $\frac{d}{dx}(\tan x)={\sec }^{2}x$
• $\frac{d}{dx}(\csc x)=-\csc x\cot x$
• $\frac{d}{dx}(\sec x)=\sec x\tan x$
• $\frac{d}{dx}(\cot x)=-{\csc }^{2}x$

A useful mnemonic: The derivative of every "co-" function (cosine, cotangent, cosecant) is negative.

Combining with Product and Quotient Rules

The real power of these new rules becomes apparent when we combine them with the differentiation rules you already know.

Example 1: Product Rule
Find the derivative of $y=x^2\sin x$.
This is a product of $f(x)=x^2$ and $g(x)=\sin x$.
Using the Product Rule, $y^{\prime }=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$:

1. $f^{\prime }(x)=2x$
2. $g^{\prime }(x)=\cos x$
3. Assemble the derivative:
   $y^{\prime }=(2x)(\sin x)+(x^2)(\cos x)=2x\sin x+x^2\cos x$

This common type of problem is worked through in the following video.

Example 2: Quotient Rule
Differentiate $f(x)=\frac{\sec x}{1+\tan x}$.
This is a quotient rule problem from your textbook (Example 2, Section 3.3).
Let the numerator be $u(x)=\sec x$ and the denominator be $v(x)=1+\tan x$.

• $u^{\prime }(x)=\sec x\tan x$
• $v^{\prime }(x)=0+{\sec }^{2}x={\sec }^{2}x$

Applying the Quotient Rule $\frac{v{u}^{\prime }-u{v}^{\prime }}{v^2}$:
$f^{\prime }(x)=\frac{(1+\tan x)(\sec x\tan x)-(\sec x)({\sec }^{2}x)}{(1+\tan x{)}^2}$
Now, simplify the numerator:
$f^{\prime }(x)=\frac{\sec x\tan x+\sec x{\tan }^{2}x-{\sec }^{3}x}{(1+\tan x{)}^2}$
Factor out $\sec x$:
$f^{\prime }(x)=\frac{\sec x(\tan x+{\tan }^{2}x-{\sec }^{2}x)}{(1+\tan x{)}^2}$
Using the identity ${\tan }^{2}x+1={\sec }^{2}x$, we can substitute ${\tan }^{2}x-{\sec }^{2}x=-1$:
$f^{\prime }(x)=\frac{\sec x(\tan x-1)}{(1+\tan x{)}^2}$
This example shows that after applying the calculus rules, trigonometric identities often play a key role in simplification.

For more practice, Professor Leonard works through another excellent Quotient Rule problem and then uses the result to find the equation of a tangent line, connecting back to earlier concepts.

Conclusion

In this lesson, we have significantly expanded your differentiation toolkit by adding the derivatives of all six basic trigonometric functions. You have seen how the derivatives of sine and cosine are derived from the limit definition and how the other four can be found by cleverly applying the Quotient Rule.

Key Takeaways:

• The derivatives of sine and cosine are fundamental: $\frac{d}{dx}(\sin x)=\cos x$ and $\frac{d}{dx}(\cos x)=-\sin x$.
• The derivatives of the other four trigonometric functions ($\tan x,\cot x,\sec x,\csc x$) can be derived from sine and cosine using the Quotient Rule.
• The six trigonometric derivative formulas should be memorized. Remember that the derivatives of the "co-" functions are negative.
• Real-world problems often require combining these new rules with the Product and Quotient Rules, followed by algebraic simplification using trigonometric identities.

To practice these skills, I recommend working through exercises 1-23 (odd numbers) from Section 3.3 of your textbook (LINK).

In our next lesson, we will introduce the Chain Rule. This is arguably the most important differentiation rule, as it will allow us to differentiate composite functions—functions within functions—such as $y=\sin (x^2)$ or $y=\sqrt{\cos (x)}$. Mastering the Chain Rule will unlock a vast new range of functions that we can analyze.