Find the 30th derivative of y = cos(3x) by finding the first few derivatives and observing the pattern that occurs.
The Correct Answer and Explanation is:
We are asked to find the 30th derivative of
y = cos(3x)
by first computing a few derivatives and observing the pattern.
Step 1: Compute the first few derivatives
1st derivative:
y′ = d/dx [cos(3x)] = -3 sin(3x)
2nd derivative:
y″ = d/dx [-3 sin(3x)] = -9 cos(3x)
3rd derivative:
y‴ = d/dx [-9 cos(3x)] = 27 sin(3x)
4th derivative:
y⁽⁴⁾ = d/dx [27 sin(3x)] = 81 cos(3x)
Step 2: Identify the pattern
The derivatives repeat every four steps, but the coefficients change by a factor of 3 each time:
- y = cos(3x)
- y′ = -3 sin(3x)
- y″ = -9 cos(3x)
- y‴ = 27 sin(3x)
- y⁽⁴⁾ = 81 cos(3x)
This shows a cyclic pattern in function type:
cos → sin → -cos → -sin → cos → …
The function alternates between sine and cosine, and the sign follows a pattern as well.
Also, the coefficient follows a geometric pattern:
Each time the derivative is taken, it is multiplied by 3.
We can summarize:
- nth derivative of cos(3x) will be:
y⁽ⁿ⁾ = 3ⁿ × cos(3x), or sin(3x), or -cos(3x), or -sin(3x) depending on n mod 4.
Step 3: Apply this to the 30th derivative
Find 30 mod 4 = 2, so it follows the pattern of the 2nd derivative, which was:
y″ = -9 cos(3x)
So the 30th derivative is:
y⁽³⁰⁾ = -3³⁰⁻² × cos(3x) = -3²⁸ × cos(3x)
Final Answer:
y⁽³⁰⁾ = -3²⁸ × cos(3x)
Explanation
To find the 30th derivative of y = cos(3x), we begin by taking the first few derivatives and identifying any repeating behavior. The first derivative gives -3 sin(3x), showing that each differentiation introduces a factor of 3 and swaps the trigonometric function between cosine and sine. Continuing this process, the second derivative becomes -9 cos(3x), the third becomes 27 sin(3x), and the fourth becomes 81 cos(3x). The pattern of functions repeats every four derivatives, forming a cycle: cos(3x), -sin(3x), -cos(3x), sin(3x), and then back to cos(3x).
Additionally, each derivative multiplies the coefficient by another 3, forming a power of 3. This suggests that the nth derivative includes the factor 3ⁿ. However, because of the cyclic nature of the trigonometric functions and signs, we also need to consider the position within the four-step cycle. The 30th derivative falls in the same place as the second derivative, because 30 mod 4 equals 2. The second derivative was -9 cos(3x), or -3² × cos(3x). So, following the same pattern, the 30th derivative should be -3²⁸ × cos(3x), because the coefficient follows 3 raised to the power of n, and the sign matches the second derivative’s pattern.
This method allows us to avoid computing all 30 derivatives individually. By recognizing the repeating structure and calculating only the first few, we simplify the problem. Recognizing patterns like this is a powerful tool in calculus, especially when dealing with trigonometric functions and higher-order derivatives.
