Derivatives of Trig Functions: Videos & Practice Problems

Understanding the derivatives of trigonometric functions is essential in calculus. The derivative of sine is $\cos \left(x\right)$, and the derivative of cosine is $-\sin \left(x\right)$. For tangent, the derivative is $\sec \left(x\right)^2$. Applying the sum and product rules allows for efficient calculation of derivatives in complex functions, reinforcing the importance of memorizing these derivatives for problem-solving.

Derivatives of Sine & Cosine

When taking derivatives of functions that include trigonometric functions, it's essential to know the specific rules that apply to sine and cosine. The derivative of the sine function, denoted as sin(x), is cos(x), while the derivative of the cosine function, cos(x), is -sin(x). Understanding these derivatives can be enhanced by visualizing them on a graph, where the derivative represents the slope of the tangent line at any given point.

For instance, at x = π/2, the slope of the tangent line to the sine function is zero, which corresponds to the value of the cosine function at that point also being zero. Conversely, at x = 2π, the slope of the tangent line to the sine function is positive, matching the positive value of the cosine function. This relationship helps reinforce why the derivatives of sine and cosine are defined as they are.

To apply these rules effectively, consider the sum rule for derivatives, which states that the derivative of a sum of functions is the sum of their derivatives. For example, if we have a function f(x) = 3x + cos(x), we can find its derivative f'(x) by differentiating each term individually. The derivative of 3x is 3, and the derivative of cos(x) is -sin(x). Thus, the final derivative is f'(x) = 3 - sin(x).

When dealing with products of functions, the product rule is necessary. For a function like f(x) = x^2 * sin(x), the product rule states that the derivative is given by left d right + right d left. Here, we differentiate the first function x^2 and multiply it by the derivative of the second function sin(x), which is cos(x). Then, we add the product of the second function sin(x) and the derivative of the first function 2x. This results in the derivative f'(x) = x^2 * cos(x) + 2x * sin(x).

By mastering these rules and practicing their application, you will enhance your ability to differentiate a wide range of functions involving trigonometric components.

Find the derivative of the function.
$h(t)=\sin\left(t\right)\cos\left(t\right)$

• A. $0$
• B. $1$
• C. $2\sin\left(t\right)\cos\left(t\right)$
• D. $-\sin^2\left(t\right)+\cos^2\left(t\right)$

Answer: (D)

Find the derivative of the function.
$f\left(x\right)=\frac{5\cos x}{2x^3}$

• A. $\frac{-5\left(x\sin x+3\cos x\right)}{2x^3}$
• B. $\frac{-5\left(x\sin x+3\cos x\right)}{2x^4}$
• C. $\frac{5x^2\cos x-10x^3\sin x}{4x^6}$
• D. $\frac{-5\sin x}{6x^2}$

Answer: (B)

Find the derivative of the function.
$y=\frac{\sin\theta}{2+\cos\theta}$

• A. $-\frac{\cos\theta}{\sin\theta}$
• B. $\frac{\cos\theta}{\left(2+\cos\theta\right)^2}$
• C. $\frac{2\cos\theta+1}{\left(2+\cos\theta\right)^2}$
• D. $\frac{2\cos\theta+\cos^2\theta-\sin^2\theta}{\left(2+\cos\theta\right)^2}$

Answer: (C)

Derivatives of Other Trig Functions

Understanding the derivatives of trigonometric functions is essential for calculus, as these derivatives form the foundation for more complex applications. The derivatives of sine and cosine functions are foundational, and it is important to extend this knowledge to other trigonometric functions such as tangent, cotangent, secant, and cosecant.

The derivative of the tangent function, tan(x), is given by:

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

This relationship can be derived by rewriting tangent in terms of sine and cosine:

\(\tan(x) = \frac{\sin(x)}{\cos(x)}\)

Using the quotient rule, which states that for two functions u and v, the derivative is:

\(\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}\)

we can find the derivative of tangent. Applying the quotient rule yields:

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

Utilizing the Pythagorean identity, sin²(x) + cos²(x) = 1, simplifies this to:

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

Similarly, the derivatives of the other trigonometric functions can be derived. For instance, the derivative of cotangent is:

\(\frac{d}{dx}(\cot(x)) = -\csc^2(x)\)

And for secant:

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

Lastly, the derivative of cosecant is:

\(\frac{d}{dx}(\csc(x)) = -\csc(x) \cot(x)\)

When applying these derivatives in practice, it is useful to remember the rules of differentiation, such as the sum rule and product rule. For example, if you need to find the derivative of a function like f(x) = 3x² + cot(x), you would apply the sum rule:

\(f'(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(\cot(x)) = 6x - \csc^2(x)\)

For a product of functions, such as f(x) = 4x \cdot sec(x), the product rule is applied:

\(f'(x) = 4x \cdot \frac{d}{dx}(\sec(x)) + \sec(x) \cdot \frac{d}{dx}(4x)\)

Substituting the derivatives gives:

\(f'(x) = 4x \cdot (\sec(x) \tan(x)) + \sec(x) \cdot 4\)

Factoring out common terms results in:

\(f'(x) = 4 \sec(x) (x \tan(x) + 1)\)

Memorizing these derivatives and practicing their application will greatly enhance your proficiency in calculus, especially when dealing with more complex functions and their derivatives.

Find the derivative of the function.
$r=\csc x-3\sin x+\tan x$

• A. $\csc x\cot x+\cos x+\sec^2x$
• B. $-\csc x\cot x-\cos x+\sec^2x$
• C. $-\csc x\cot x-3\cos x+\sec^2x$
• D. $-\cot x-3\cos x+\sec^2x$

Answer: (C)

Find the derivative of the function.
$f\left(x\right)=4x^2\sec x-\sqrt{x}$

• A. $8x\sec x+\frac{1}{2\sqrt{x}}$
• B. $8x\sec x\tan x-\frac{1}{2\sqrt{x}}$
• C. $4x^2\sec x\tan x+8x\sec x-\frac12\sqrt{x}$
• D. $4x^2\sec x\tan x+8x\sec x-\frac{1}{2\sqrt{x}}$

Answer: (D)

Find the derivative of the function.
$y=\frac{\cot\theta}{3+\sec\theta}$

• A. $\frac{-\csc^2\theta\left(3+\sec\theta\right)-\sec\theta}{\left(3+\sec\theta\right)^2}$
• B. $\frac{\sec\theta+\csc^2\theta\left(3+\sec\theta\right)}{\left(3+\sec\theta\right)^2}$
• C. $\frac{1}{\left(3+\sec\theta\right)^2}$
• D. $\frac{-\csc^2\theta}{\sec\theta\tan\theta}$

Answer: (A)

Derivatives of Other Trig Functions Example 1

To find the slope of the tangent line for the function \( f(x) = \sec(x) \) at the point \( x = \frac{\pi}{4} \), we first need to determine the derivative of the function, denoted as \( f'(x) \). The derivative of the secant function is given by the formula:

\[ f'(x) = \sec(x) \tan(x) \]

Next, we evaluate the derivative at the specific point \( x = \frac{\pi}{4} \). This involves substituting \( \frac{\pi}{4} \) into the derivative:

\[ f'\left(\frac{\pi}{4}\right) = \sec\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) \]

We know from trigonometric values that:

• \( \tan\left(\frac{\pi}{4}\right) = 1 \)
• \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \)

Substituting these values into the derivative gives us:

\[ f'\left(\frac{\pi}{4}\right) = \sqrt{2} \cdot 1 = \sqrt{2} \]

Thus, the slope of the tangent line to the function \( f(x) = \sec(x) \) at \( x = \frac{\pi}{4} \) is \( \sqrt{2} \). This demonstrates the application of derivative rules for trigonometric functions, reinforcing the concept that the derivative provides the slope of the tangent line at any given point on the curve.

Example 2: Chain Rule

To find the derivatives of the functions \( f(x) = \sin^5(x) \) and \( g(x) = \sin(x^5) \), we can apply the chain rule, which is essential for differentiating composite functions.

Starting with the first function, \( f(x) = \sin^5(x) \), we can rewrite it as \( f(x) = (\sin(x))^5 \). Here, the outer function is \( u^5 \) where \( u = \sin(x) \). According to the chain rule, the derivative \( f'(x) \) is given by:

\[f'(x) = 5(\sin(x))^4 \cdot \frac{d}{dx}(\sin(x))\]

Since the derivative of \( \sin(x) \) is \( \cos(x) \), we substitute this into our equation:

\[f'(x) = 5(\sin(x))^4 \cdot \cos(x)\]

Thus, the derivative of \( f(x) = \sin^5(x) \) is:

\[f'(x) = 5 \cos(x) \sin^4(x)\]

Next, we consider the second function, \( g(x) = \sin(x^5) \). Here, the outer function is \( \sin(u) \) where \( u = x^5 \). Again applying the chain rule, we find:

\[g'(x) = \cos(x^5) \cdot \frac{d}{dx}(x^5)\]

Using the power rule, the derivative of \( x^5 \) is \( 5x^4 \). Therefore, we can express \( g'(x) \) as:

\[g'(x) = \cos(x^5) \cdot 5x^4\]

Rearranging this gives us the final form of the derivative:

\[g'(x) = 5x^4 \cos(x^5)\]

In summary, the derivatives of the functions are:

\[f'(x) = 5 \cos(x) \sin^4(x)\]

and

\[g'(x) = 5x^4 \cos(x^5)\]

Practicing these derivatives will enhance your understanding of the chain rule, especially when dealing with trigonometric functions.

Find the derivative of the function.
$f(x)=\sin(3x^2)$

• A. $\cos\left(6x\right)$
• B. $-6x\cos\left(3x^2\right)$
• C. $6x\cos\left(3x^2\right)$
• D. $\cos\left(3x^2\right)$

Answer: (C)

Find the derivative of the function.
$y=3\cos^4\theta$

• A. $-4\sin^3\theta$
• B. $-12\sin^3\theta$
• C. $4\sin\theta\cos^3\theta$
• D. $-12\sin\theta\cos^3\theta$

Answer: (D)

Find the derivative of the function.
$f\left(t\right)=\sec\left(4t+5\right)$

• A. $4\sec\left(4t+5\right)$
• B. $4\sec\left(4t+5\right)\tan\left(4t+5\right)$
• C. $4\sec\left(t\right)\tan\left(t\right)$
• D. $\sec\left(4\right)\tan\left(4\right)$

Answer: (B)

Here’s what students ask on this topic:

What is the derivative of sin(x) and how can it be applied?

The derivative of sin(x) is cos(x). This means that the slope of the tangent line to the graph of sin(x) at any point is equal to the value of cos(x) at that same point. For example, if you have a function f(x) = 3sin(x), you can find its derivative using the constant multiple rule. The derivative would be f'(x) = 3cos(x). This concept is widely used in physics, engineering, and other fields to model oscillatory motion, such as waves or vibrations. Memorizing this derivative is essential for solving problems efficiently in calculus.

How do you find the derivative of cos(x)?

The derivative of cos(x) is -sin(x). This means that the slope of the tangent line to the graph of cos(x) at any point is equal to the negative value of sin(x) at that same point. For example, if you have a function f(x) = 2cos(x), you can find its derivative using the constant multiple rule. The derivative would be f'(x) = -2sin(x). This derivative is particularly useful in applications involving periodic motion, such as analyzing electrical signals or mechanical vibrations.

What is the derivative of tan(x) and how is it derived?

The derivative of tan(x) is sec²(x). To derive this, you can rewrite tan(x) as sin(x)/cos(x) and apply the quotient rule. The quotient rule states that the derivative of a function f(x) = g(x)/h(x) is given by (low d high - high d low) / (low²). Applying this to tan(x), you get:

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

Using the Pythagorean identity sin²(x) + cos²(x) = 1, this simplifies to sec²(x). Memorizing this derivative is crucial for solving problems involving tangent functions efficiently.

How do you use the product rule to find the derivative of a function involving trig functions?

The product rule states that the derivative of a product of two functions, f(x) = u(x)v(x), is given by u'(x)v(x) + u(x)v'(x). For example, if f(x) = x²sin(x), you can apply the product rule as follows:

$\left(x^2\right)\text{'}\left(\sin x\right)+\left(x^2\right)\left(\cos x\right)$

This simplifies to 2xsin(x) + x²cos(x). The product rule is essential for finding derivatives of functions where trig functions are multiplied by polynomials or other functions.

What are the derivatives of the remaining trig functions (cotangent, secant, and cosecant)?

The derivatives of the remaining trig functions are as follows:

• The derivative of cot(x) is -csc²(x).
• The derivative of sec(x) is sec(x)tan(x).
• The derivative of csc(x) is -csc(x)cot(x).

These derivatives can be derived using the quotient rule by expressing each function in terms of sin(x) and cos(x). For example, cot(x) = cos(x)/sin(x), and applying the quotient rule yields -csc²(x). Memorizing these derivatives is highly recommended for efficiency, as they are frequently used in calculus problems involving trigonometric functions.