Derivatives of Trig Functions: Videos & Practice Problems

The derivatives of sine and cosine functions are foundational in calculus, with the derivative of sine being $d\left(\sin \right(x\left)\right)=\cos \left(x\right)$ and cosine yielding $d\left(\cos \right(x\left)\right)=-\sin \left(x\right)$. Higher order derivatives follow a cyclical pattern every four derivatives. Additionally, derivatives of other trigonometric functions can be derived using the quotient rule, with the tangent's derivative being $d\left(\tan \right(x\left)\right)=\sec \left(x\right)^2$.

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 the slope of the tangent lines to the graphs of these functions.

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 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), the derivative can be calculated as follows:

f'(x) = d/dx(3x) + d/dx(cos(x)) = 3 - sin(x)

Thus, the final derivative is f'(x) = 3 - sin(x).

When dealing with products of functions, the product rule is necessary. The product rule states that for two functions u(x) and v(x), the derivative is given by:

f'(x) = u'v + uv'

For example, if we have f(x) = x^2 * sin(x), we apply the product rule:

f'(x) = (d/dx(x^2)) * sin(x) + x^2 * (d/dx(sin(x)))

Calculating each derivative, we find:

f'(x) = 2x * sin(x) + x^2 * cos(x)

Rearranging gives us the final derivative:

f'(x) = x^2 * cos(x) + 2x * sin(x)

With these foundational rules for derivatives of trigonometric functions, you can confidently tackle more complex derivatives involving sine and cosine, enhancing your calculus skills.

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

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

Answer: (C)

Higher Order Derivatives of Sine & Cosine

Understanding how to find higher order derivatives of sine and cosine functions can significantly simplify your calculations. Instead of performing repeated differentiation, you can utilize a pattern that emerges from the derivatives of these trigonometric functions.

When you differentiate the sine function, the first few derivatives are:

• First derivative: \( \frac{d}{dx}(\sin x) = \cos x \)
• Second derivative: \( \frac{d^2}{dx^2}(\sin x) = -\sin x \)
• Third derivative: \( \frac{d^3}{dx^3}(\sin x) = -\cos x \)
• Fourth derivative: \( \frac{d^4}{dx^4}(\sin x) = \sin x \)

This cycle repeats every four derivatives. Therefore, to find the \( n \)-th derivative of sine, you can divide \( n \) by 4 and use the remainder to determine which derivative corresponds to the original function:

For example, to find the 37th derivative of sine:

1. Divide 37 by 4, which gives a quotient of 9 and a remainder of 1.

2. The 37th derivative corresponds to the first derivative of sine, which is \( \cos x \).

Similarly, for the cosine function, the derivatives are:

• First derivative: \( \frac{d}{dx}(\cos x) = -\sin x \)
• Second derivative: \( \frac{d^2}{dx^2}(\cos x) = -\cos x \)
• Third derivative: \( \frac{d^3}{dx^3}(\cos x) = \sin x \)
• Fourth derivative: \( \frac{d^4}{dx^4}(\cos x) = \cos x \)

To find the 58th derivative of cosine:

1. Divide 58 by 4, yielding a quotient of 14 and a remainder of 2.

2. The 58th derivative corresponds to the second derivative of cosine, which is \( -\cos x \).

In summary, for any higher order derivative of sine or cosine, simply divide the order by 4 and use the remainder to identify the corresponding derivative from the established cycle. This method allows for quick and efficient calculations without the need for extensive differentiation.

Find the indicated derivative.
$f^{19}\left(x\right)$ of $f\left(x\right)=\cos x$

• A. $\cos x$
• B. $\sin x$
• C. $-\cos x$
• D. $-\sin x$

Answer: (B)

Find the indicated derivative.
$d^{1002}/d{x}^{1002}\left(\sin x\right)$

• A. $\cos x$
• B. $\sin x$
• C. $-\cos x$
• D. $-\sin x$

Answer: (D)

Find the indicated derivative.
$d^{52}y/d{x}^{52}$ of $y=\cos x$

• A. $\cos x$
• B. $\sin x$
• C. $-\cos x$
• D. $-\sin x$

Answer: (A)

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 fundamental, but there are additional trigonometric functions that also require attention. Each of these functions has a derivative that can be expressed in terms of other trigonometric functions.

For instance, the derivative of the tangent function, denoted as \( \tan(x) \), is given by:

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

To derive this, we can express tangent as the quotient of sine and cosine:

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

Using the quotient rule, which states that if \( u \) and \( v \) are functions, then:

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

we can find the derivative of tangent. Applying this rule, we have:

\[ \frac{d}{dx} \tan(x) = \frac{\cos(x) \cdot \cos(x) - \sin(x) \cdot (-\sin(x))}{\cos^2(x)} \]

which simplifies to:

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

Similarly, the derivatives of the remaining trigonometric functions can be derived. The derivatives of the cofunctions, such as cotangent, secant, and cosecant, are particularly noteworthy as they are negative:

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

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

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

When finding the derivative of a function that combines these trigonometric functions, such as \( f(x) = 3x^2 + \cot(x) \), we can apply the sum rule:

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

For products of functions, such as \( f(x) = 4x \sec(x) \), the product rule is used:

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

Calculating this gives:

\[ f'(x) = 4 \sec(x) + 4x \sec(x) \tan(x) \]

Factoring out common terms results in:

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

Memorizing these derivatives is crucial, as they will frequently appear in various calculus problems. However, if you forget any, you can always revert to expressing the functions in terms of sine and cosine and applying the quotient rule to derive them again.

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

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

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. $\frac{-{\csc }^2\theta (3+\sec \theta )-\sec \theta }{{(3+\sec \theta )}^2}$
• B. $\frac{\sec \theta +{\csc }^2\theta (3+\sec \theta )}{{(3+\sec \theta )}^2}$
• C. $\frac{1}{{(3+\sec \theta )}^2}$
• 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:

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

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 illustrates how the principles of derivatives apply to trigonometric functions, allowing us to find slopes of tangent lines effectively.

Here’s what students ask on this topic:

What is the derivative of sin(x)?

The derivative of the sine function, sin(x), is the cosine function, cos(x). In mathematical notation, this is expressed as:

$d\left(\sin \right(x\left)\right)=\cos \left(x\right)$

This means that the rate of change of the sine function at any point x is equal to the value of the cosine function at that point.

What is the derivative of cos(x)?

The derivative of the cosine function, cos(x), is the negative sine function, -sin(x). In mathematical notation, this is expressed as:

$d\left(\cos \right(x\left)\right)=-\sin \left(x\right)$

This means that the rate of change of the cosine function at any point x is equal to the negative value of the sine function at that point.

How do you find the higher order derivatives of sin(x) and cos(x)?

Higher order derivatives of sin(x) and cos(x) follow a cyclical pattern every four derivatives. For sin(x):

$d\left(\sin \right(x\left)\right)=\cos \left(x\right)$$d\left(\cos \right(x\left)\right)=-\sin \left(x\right)$$d(-\sin (x\left)\right)=-\cos \left(x\right)$$d(-\cos (x\left)\right)=\sin \left(x\right)$

For cos(x), the pattern is similar but shifted by one derivative. This cyclical pattern allows you to find higher order derivatives quickly by dividing the order number by 4 and using the remainder to determine the corresponding derivative.

What is the derivative of tan(x)?

The derivative of the tangent function, tan(x), is the secant squared function, sec²(x). In mathematical notation, this is expressed as:

$d\left(\tan \right(x\left)\right)={\sec }^2\left(x\right)$

This means that the rate of change of the tangent function at any point x is equal to the square of the secant function at that point.

How do you find the derivative of a product of functions involving trigonometric functions?

To find the derivative of a product of functions involving trigonometric functions, you use the product rule. The product rule states:

$\left(\mathrm{fg}\right)\text{'}=f\text{'}g+fg\text{'}$

For example, if you have f(x) = x²sin(x), the derivative f'(x) is found as follows:

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

This simplifies to:

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

So, the derivative of x²sin(x) is x²cos(x) + 2xsin(x).