Last updated on July 22nd, 2025
We use the derivative of cos(x^3), which involves applying the chain rule, as a tool to measure how the function changes in response to a slight change in x. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of cos(x^3) in detail.
We now understand the derivative of cos(x^3). It is commonly represented as d/dx (cos(x^3)) or (cos(x^3))', and its value is -3x²sin(x^3).
The function cos(x^3) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Cosine Function: cos(x^3) involves the composition of functions.
Chain Rule: Rule for differentiating cos(x^3) due to its composite nature.
Sine Function: sin(x) is the derivative of cos(x).
The derivative of cos(x^3) can be denoted as d/dx (cos(x^3)) or (cos(x^3))'. The formula we use to differentiate cos(x^3) is: d/dx (cos(x^3)) = -3x²sin(x^3)
formula applies to all x where x is within the domain of the function.
We can derive the derivative of cos(x^3) using proofs. To show this, we will use the trigonometric identities along with the rules of differentiation. There are several methods we use to prove this, such aBy Chain Rule
We will now demonstrate that the differentiation of cos(x^3) results in -3x²sin(x^3) using the chain rule:
To prove the differentiation of cos(x^3) using the chain rule, We use the formula: cos(x^3) = cos(u), where u = x^3
The derivative of cos(u) is -sin(u), and the derivative of u = x^3 is 3x².
By chain rule: d/dx [cos(u)] = -sin(u) * du/dx
Let’s substitute u = x^3, d/dx (cos(x^3)) = -sin(x^3) * 3x²
Hence, d/dx (cos(x^3)) = -3x²sin(x^3).
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 cos(x^3).
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 cos(x^3), 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=0, the derivative of cos(x^3) = -3x²sin(x^3) = 0 because sin(0) = 0.
When x is not a real number, the derivative is undefined because x^3 must be a real number for cos(x^3) to be defined.
Students frequently make mistakes when differentiating cos(x^3). 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 cos(x^3)·x²
Here, we have f(x) = cos(x^3)·x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = cos(x^3) and v = x².
Let’s differentiate each term, u′ = d/dx (cos(x^3)) = -3x²sin(x^3) v′ = d/dx (x²) = 2x
Substituting into the given equation, f'(x) = (-3x²sin(x^3))·x² + cos(x^3)·2x
Let’s simplify terms to get the final answer, f'(x) = -3x⁴sin(x^3) + 2xcos(x^3)
Thus, the derivative of the specified function is -3x⁴sin(x^3) + 2xcos(x^3).
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.
A company is analyzing the cost function represented by y = cos(x^3), where y represents the cost at a production level x. If x = 1 meter, determine the rate of change of the cost.
We have y = cos(x^3) (cost function)...(1)
Now, we will differentiate the equation (1)
Take the derivative of cos(x^3): dy/dx = -3x²sin(x^3)
Given x = 1 (substitute this into the derivative) dy/dx = -3(1)²sin(1^3) dy/dx = -3sin(1)
Hence, the rate of change of the cost at x = 1 is -3sin(1).
We find the rate of change of the cost at x = 1, which provides insight into how the cost function behaves at that particular production level.
Derive the second derivative of the function y = cos(x^3).
The first step is to find the first derivative, dy/dx = -3x²sin(x^3)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-3x²sin(x^3)]
Here we use the product rule, d²y/dx² = -3[d/dx(x²)sin(x^3) + x²d/dx(sin(x^3))] d²y/dx² = -3[2xsin(x^3) + x²(3x²cos(x^3))] d²y/dx² = -3[2xsin(x^3) + 3x⁴cos(x^3)]
Therefore, the second derivative of the function y = cos(x^3) is -3(2xsin(x^3) + 3x⁴cos(x^3)).
We use the step-by-step process, starting with the first derivative. Using the product rule, we differentiate the components and simplify the terms to find the final answer.
Prove: d/dx (sin(x^3)) = 3x²cos(x^3).
Let’s start using the chain rule: Consider y = sin(x^3)
The derivative of sin(u) is cos(u), and the derivative of u = x^3 is 3x².
Using the chain rule: dy/dx = cos(x^3) * 3x²
Hence, d/dx (sin(x^3)) = 3x²cos(x^3).
In this step-by-step process, we used the chain rule to differentiate the equation. We replace sin(x^3) with its derivative and multiply by the derivative of the inner function to derive the equation.
Solve: d/dx (cos(x^3)/x)
To differentiate the function, we use the quotient rule: d/dx (cos(x^3)/x) = (d/dx (cos(x^3))·x - cos(x^3)·d/dx(x))/x²
We will substitute d/dx (cos(x^3)) = -3x²sin(x^3) and d/dx(x) = 1 = (-3x²sin(x^3)·x - cos(x^3)·1)/x² = (-3x³sin(x^3) - cos(x^3))/x²
Therefore, d/dx (cos(x^3)/x) = (-3x³sin(x^3) - cos(x^3))/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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