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Last updated on September 15, 2025
We use the derivative of sin³(x), which helps us understand how the function changes in response to a small change in x. Derivatives are useful for calculating various changes in real-life situations. We will now discuss the derivative of sin³(x) in detail.
We now understand the derivative of sin³(x). It is commonly represented as d/dx (sin³x) or (sin³x)', and its value is 3sin²x·cosx. The function sin³x has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:
Sine Function: (sin(x)).
Chain Rule: Rule for differentiating composite functions like sin³x.
Power Rule: Rule for differentiating functions of the form x^n.
The derivative of sin³x can be denoted as d/dx (sin³x) or (sin³x)'.
The formula we use to differentiate sin³x is: d/dx (sin³x) = 3sin²x·cosx
The formula applies to all x where sin(x) is defined.
We can derive the derivative of sin³x using proofs. To show this, we will use trigonometric identities along with differentiation rules. There are several methods we use to prove this, such as: Using the Chain Rule Using the Product Rule We will now demonstrate that the differentiation of sin³x results in 3sin²x·cosx using these methods:
To prove the differentiation of sin³x using the chain rule, We use the formula: Let y = (sin(x))³
By the chain rule, dy/dx = 3(sin(x))²·d/dx(sin(x))
Since d/dx(sin(x)) = cos(x), dy/dx = 3(sin(x))²·cos(x) Hence, proved.
We can also prove the derivative of sin³x using the product rule. Here's the step-by-step process: Let y = sin(x)·sin(x)·sin(x)
By the product rule, d/dx[u·v·w] = uv'w + u'vw + uvw' Let u = sin(x), v = sin(x), w = sin(x)
Then u' = cos(x), v' = cos(x), w' = cos(x) dy/dx = sin(x)·sin(x)·cos(x) + sin(x)·cos(x)·sin(x) + cos(x)·sin(x)·sin(x) = 3sin²(x)·cos(x)
Thus, d/dx(sin³x) = 3sin²x·cosx.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little complex.
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 help in understanding functions like sin³(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 sin³(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.
When x is 0, the derivative of sin³(x) = 3sin²(0)·cos(0), which is 0.
Students frequently make mistakes when differentiating sin³x. 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 (sin³x·cos²x)
Here, we have f(x) = sin³x·cos²x.
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin³x and v = cos²x.
Let’s differentiate each term, u′= d/dx (sin³x) = 3sin²x·cosx v′= d/dx (cos²x) = -2cosx·sinx
Substituting into the given equation, f'(x) = (3sin²x·cosx)·(cos²x) + (sin³x)·(-2cosx·sinx)
Let’s simplify terms to get the final answer, f'(x) = 3sin²x·cos³x - 2sin⁴x·cosx
Thus, the derivative of the specified function is 3sin²x·cos³x - 2sin⁴x·cosx.
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 tree is modeled by the function y = sin³(x) meters, where x is time in months. Calculate the rate of change of the tree's height when x = π/6 months.
We have y = sin³(x) (height of the tree)...(1)
Now, we will differentiate the equation (1) Take the derivative of sin³(x): dy/dx = 3sin²x·cosx
Given x = π/6, substitute this into the derivative: dy/dx = 3sin²(π/6)·cos(π/6) = 3(1/2)²·(√3/2) = 3(1/4)·(√3/2) = 3√3/8
Hence, the rate of change of the tree's height at x = π/6 is 3√3/8 meters per month.
We find the rate of change of the tree's height at x = π/6 by differentiating the function y = sin³(x) and then substituting x = π/6 into the derivative.
Derive the second derivative of the function y = sin³(x).
The first step is to find the first derivative, dy/dx = 3sin²(x)·cos(x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [3sin²(x)·cos(x)]
Here we use the product rule, d²y/dx² = 3[d/dx(sin²(x))·cos(x) + sin²(x)·d/dx(cos(x))] = 3[2sin(x)cos(x)·cos(x) - sin²(x)sin(x)] = 6sin(x)cos²(x) - 3sin³(x)
Therefore, the second derivative of the function y = sin³(x) is 6sin(x)cos²(x) - 3sin³(x).
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate 3sin²(x)·cos(x). We then simplify the terms to find the final answer.
Prove: d/dx ((sin(x))³) = 3sin²(x)cos(x).
Let’s start using the chain rule: Consider y = (sin(x))³
To differentiate, we use the chain rule: dy/dx = 3(sin(x))²·d/dx[sin(x)]
Since the derivative of sin(x) is cos(x), dy/dx = 3(sin(x))²·cos(x) Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(x) with its derivative. As a final step, we obtain the derivative of the function.
Solve: d/dx (sin³x/x)
To differentiate the function, we use the quotient rule: d/dx (sin³x/x) = (d/dx (sin³x)·x - sin³x·d/dx(x))/x²
We will substitute d/dx(sin³x) = 3sin²x·cosx and d/dx(x) = 1 = (3sin²x·cosx·x - sin³x·1) / x² = (3xsin²x·cosx - sin³x) / x²
Therefore, d/dx (sin³x/x) = (3xsin²x·cosx - sin³x) / x²
In this process, we differentiate the given function using the product rule and 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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