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109 LearnersLast updated on September 22, 2025
Derivative of sin(x^3)
We use the derivative of sin(x^3), which involves applying the chain rule, to measure how the sine function changes in response to a slight change in x. Derivatives help us calculate rates of change in real-life situations. We will now talk about the derivative of sin(x^3) in detail.
What is the Derivative of sin(x^3)?
We now understand the derivative of sin(x^3). It is commonly represented as d/dx (sin(x^3)) or (sin(x^3))', and its value is 3x²cos(x^3).
The function sin(x^3) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Sine Function: sin(x^3) is a composition of the sine function and x^3.
Chain Rule: Rule for differentiating compositions of functions.
Cosine Function: cos(x) is the derivative of sin(x).
Derivative of sin(x^3) Formula
The derivative of sin(x^3) can be denoted as d/dx (sin(x^3)) or (sin(x^3))'.
The formula we use to differentiate sin(x^3) is: d/dx (sin(x^3)) = 3x²cos(x^3)
The formula applies to all x in the domain of sin(x^3).
Proofs of the Derivative of sin(x^3)
We can derive the derivative of sin(x^3) using proofs. To show this, we will use the chain rule of differentiation. Here is how it is done:
Using Chain Rule The derivative of sin(x^3) can be found using the chain rule.
Consider u(x) = x^3, then f(u) = sin(u). By the chain rule, d/dx [f(u(x))] = f'(u(x))·u'(x).
We know that f'(u) = cos(u) and u'(x) = 3x².
Therefore, the derivative is: d/dx (sin(x^3)) = cos(x^3)·3x² = 3x²cos(x^3), hence proved.
Higher-Order Derivatives of sin(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 sin(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 sin(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).
Special Cases:
The derivative is undefined for x where the function is not differentiable or where x^3 causes the sine function to have undefined behavior, such as complex numbers in certain contexts. When x = 0, the derivative of sin(x^3) = 3(0)²cos(0) = 0.
Common Mistakes and How to Avoid Them in Derivatives of sin(x^3)
Students frequently make mistakes when differentiating sin(x^3). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Mistake 1
Not applying the Chain Rule correctly
Students may forget to apply the chain rule, which can lead to incorrect results. Always remember to differentiate the outer function and multiply by the derivative of the inner function.
Mistake 2
Forgetting the power rule
They might not remember that when differentiating x^3, the power rule needs to be applied. The derivative of x^3 is 3x². Ensure this is calculated and included in the chain rule application.
Mistake 3
Incorrect use of trigonometric identities
While differentiating functions like sin(x^3), students might misapply trigonometric identities. Always ensure to differentiate using the correct identity, e.g., derivative of sin(x) is cos(x).
Mistake 4
Not writing Constants and Coefficients
There is a common mistake that students sometimes forget to multiply the coefficients before the derivative, such as the 3 from x^3. Ensure that these coefficients are included in the final derivative.
Mistake 5
Not simplifying the equation
Students often skip steps and directly arrive at results without simplifying the equation, which can lead to errors. Write each step clearly and simplify the expressions to avoid mistakes.
Examples Using the Derivative of sin(x^3)
Problem 1
Calculate the derivative of sin(x^3)·x²
Here, we have f(x) = sin(x^3)·x².
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(x^3) and v = x².
Let’s differentiate each term, u′ = d/dx (sin(x^3)) = 3x²cos(x^3) v′ = d/dx (x²) = 2x
Substituting into the given equation, f'(x) = (3x²cos(x^3))·x² + (sin(x^3))·(2x)
Let’s simplify terms to get the final answer, f'(x) = 3x⁴cos(x^3) + 2xsin(x^3)
Thus, the derivative of the specified function is 3x⁴cos(x^3) + 2xsin(x^3).
Explanation
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.
Problem 2
A balloon rises such that its height y is given by y = sin(x^3) where x represents time in seconds. If x = 1 second, determine the rate of change of height at that time.
We have y = sin(x^3) (height of the balloon)...(1)
Now, we will differentiate the equation (1)
Take the derivative sin(x^3): dy/dx = 3x²cos(x^3) Given x = 1, substitute this into the derivative: dy/dx = 3(1)²cos(1^3) = 3cos(1)
Hence, we get the rate of change of height at x = 1 second as 3cos(1).
Explanation
We find the rate of change of the balloon's height at x = 1 second, which indicates how fast the height is increasing at that specific moment.
Problem 3
Derive the second derivative of the function y = sin(x^3).
The first step is to find the first derivative, dy/dx = 3x²cos(x^3)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [3x²cos(x^3)]
Here we use the product rule, d²y/dx² = 3[d/dx (x²)cos(x^3) + x²d/dx (cos(x^3))] = 3[2xcos(x^3) - x²·3x²sin(x^3)] = 6xcos(x^3) - 9x⁴sin(x^3)
Therefore, the second derivative of the function y = sin(x^3) is 6xcos(x^3) - 9x⁴sin(x^3).
Explanation
We use a step-by-step process, where we start with the first derivative. Using the product rule, we differentiate 3x²cos(x^3). We then substitute the identity and simplify the terms to find the final answer.
Problem 4
Prove: d/dx (sin²(x^3)) = 2sin(x^3)cos(x^3)·3x².
Let’s start using the chain rule: Consider y = sin²(x^3) [sin(x^3)]²
To differentiate, we use the chain rule: dy/dx = 2sin(x^3)·d/dx [sin(x^3)] Since the derivative of sin(x^3) is 3x²cos(x^3), dy/dx = 2sin(x^3)·3x²cos(x^3)
Substituting y = sin²(x^3), d/dx (sin²(x^3)) = 2sin(x^3)·3x²cos(x^3)
Hence proved.
Explanation
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(x^3) with its derivative. As a final step, we substitute y = sin²(x^3) to derive the equation.
Problem 5
Solve: d/dx (sin(x^3)/x)
To differentiate the function, we use the quotient rule: d/dx (sin(x^3)/x) = (d/dx (sin(x^3))·x - sin(x^3)·d/dx(x))/x²
We will substitute d/dx (sin(x^3)) = 3x²cos(x^3) and d/dx(x) = 1 = (3x²cos(x^3)·x - sin(x^3)·1)/x² = (3x³cos(x^3) - sin(x^3))/x²
Therefore, d/dx (sin(x^3)/x) = (3x³cos(x^3) - sin(x^3))/x²
Explanation
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.
FAQs on the Derivative of sin(x^3)
1.Find the derivative of sin(x^3).
2.Can we use the derivative of sin(x^3) in real life?
3.Is it possible to take the derivative of sin(x^3) at the point where x = 0?
4.What rule is used to differentiate sin(x^3)/x?
5.Are the derivatives of sin(x^3) and sin^(-1)(x^3) the same?
Important Glossaries for the Derivative of sin(x^3)
- Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
- Chain Rule: A rule in calculus for differentiating compositions of functions.
- Sine Function: A trigonometric function, often represented as sin(x), that describes a smooth periodic oscillation.
- Cosine Function: A trigonometric function, often represented as cos(x), which is the derivative of the sine function.
- Product Rule: A rule for finding the derivative of the product of two functions.
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Jaskaran Singh Saluja
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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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