Last updated on July 16th, 2025
We use the derivative of 3cos(x), which is -3sin(x), to understand how the cosine function changes in response to a slight change in x. Derivatives help us calculate various changes in real-life situations. We will now discuss the derivative of 3cos(x) in detail.
We now understand the derivative of 3cos(x). It is commonly represented as d/dx (3cos x) or (3cos x)', and its value is -3sin(x). The function 3cos(x) has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Cosine Function: cos(x)
Constant Multiple Rule: Rule for differentiating a constant times a function.
Sine Function: sin(x)
The derivative of 3cos(x) can be denoted as d/dx (3cos x) or (3cos x)'.
The formula we use to differentiate 3cos(x) is: d/dx (3cos x) = -3sin x
The formula applies to all x where the function is defined.
We can derive the derivative of 3cos(x) using proofs. To show this, we will use trigonometric identities along with the rules of differentiation.
There are several methods we use to prove this, such as:
We will now demonstrate that the differentiation of 3cos(x) results in -3sin(x) using these methods:
We use the formula: d/dx [a·f(x)] = a·f'(x) Let a = 3 and f(x) = cos(x) f'(x) = d/dx (cos x) = -sin x Thus, d/dx (3cos x) = 3·(-sin x) = -3sin x
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky.
To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like 3cos(x).
For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(x). Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.
For the nth Derivative of 3cos(x), we generally use fⁿ(x) for the nth derivative of a function f(x) which tells us the change in the rate of change, continuing for higher-order derivatives.
When x is π/2, the derivative is -3sin(π/2), which is -3. When x is 0, the derivative of 3cos x = -3sin(0), which is 0.
Students frequently make mistakes when differentiating 3cos(x). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of 3cos(x)·e^x
Here, we have f(x) = 3cos(x)·e^x.
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 3cos(x) and v = e^x.
Let’s differentiate each term, u′ = d/dx (3cos x) = -3sin x v′ = d/dx (e^x) = e^x Substituting into the given equation, f'(x) = (-3sin x)·(e^x) + (3cos x)·(e^x)
Let’s simplify terms to get the final answer, f'(x) = e^x(-3sin x + 3cos x)
Thus, the derivative of the specified function is e^x(-3sin x + 3cos x).
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.
The height of a Ferris wheel is represented by the function y = 3cos(x), where y is the height of a specific point at an angle x. If x = π/3 radians, determine the rate at which the height changes.
We have y = 3cos(x) (height of the Ferris wheel)...(1)
Now, we will differentiate the equation (1). Take the derivative of 3cos(x): dy/dx = -3sin(x) Given x = π/3 (substitute this into the derivative)
dy/dx = -3sin(π/3) = -3(√3/2) = -3√3/2
Hence, the rate of change of height at x = π/3 is -3√3/2.
We find the rate of change of height at x = π/3 by differentiating the height function. This shows how quickly the height of the Ferris wheel changes at a certain angle.
Derive the second derivative of the function y = 3cos(x).
The first step is to find the first derivative, dy/dx = -3sin(x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-3sin(x)] d²y/dx² = -3cos(x)
Therefore, the second derivative of the function y = 3cos(x) is -3cos(x).
We use the step-by-step process, where we start with the first derivative. Then we differentiate -3sin(x) to find the second derivative, resulting in -3cos(x).
Prove: d/dx (3cos²x) = -6cos(x)sin(x).
Let’s start using the chain rule: Consider y = 3cos²(x) = 3[cos(x)]²
To differentiate, we use the chain rule: dy/dx = 3·2cos(x)·d/dx [cos(x)]
Since the derivative of cos(x) is -sin(x), dy/dx = 6cos(x)(-sin(x)) = -6cos(x)sin(x)
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace cos(x) with its derivative. As a final step, we simplify to derive the equation.
Solve: d/dx (3cos x/x)
To differentiate the function, we use the quotient rule: d/dx (3cos x/x) = (d/dx (3cos x)·x - 3cos x·d/dx(x))/x²
We will substitute d/dx (3cos x) = -3sin x and d/dx (x) = 1 = (-3sin x·x - 3cos x·1)/x² = (-3xsin x - 3cos x)/x²
Therefore, d/dx (3cos x/x) = (-3xsin x - 3cos x)/x²
In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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