Derivative of 9sinx
2026-02-28 11:20 Diff

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Last updated on August 5, 2025

We use the derivative of 9sin(x), which is 9cos(x), as a tool to measure how the sine function changes in response to a slight change in x. Derivatives help us calculate changes in various contexts, such as physics and engineering. We will now discuss the derivative of 9sin(x) in detail.

What is the Derivative of 9sinx?

We now understand the derivative of 9sinx. It is commonly represented as d/dx (9sinx) or (9sinx)', and its value is 9cosx. The function 9sinx has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below: - Sine Function: sin(x) is a fundamental trigonometric function. - Constant Multiple Rule: Rule for differentiating functions multiplied by a constant. - Cosine Function: cos(x) is another fundamental trigonometric function.

Derivative of 9sinx Formula

The derivative of 9sinx can be denoted as d/dx (9sinx) or (9sinx)'. The formula we use to differentiate 9sinx is: d/dx (9sinx) = 9cosx The formula applies to all x.

Proofs of the Derivative of 9sinx

We can derive the derivative of 9sinx 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 as: - By First Principle - Using Chain Rule We will now demonstrate that the differentiation of 9sinx results in 9cosx using the above-mentioned methods: By First Principle The derivative of 9sinx can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of 9sinx using the first principle, we will consider f(x) = 9sinx. Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1) Given that f(x) = 9sinx, we write f(x + h) = 9sin(x + h). Substituting these into equation (1), f'(x) = limₕ→₀ [9sin(x + h) - 9sinx] / h = 9 * limₕ→₀ [sin(x + h) - sinx] / h = 9 * limₕ→₀ [2cos((x + h + x)/2)sin(h/2)] / h = 9 * limₕ→₀ [cos(x + h/2)sin(h/2)] / (h/2) Using limit formulas, limₕ→₀ (sin(h/2))/(h/2) = 1. f'(x) = 9cosx Hence, proved. Using Chain Rule To prove the differentiation of 9sinx using the chain rule, We use the formula: Let u = sinx, then f(x) = 9u The derivative becomes d/dx (9u) = 9 * d/dx (u) And d/dx (sinx) = cosx, So d/dx (9sinx) = 9cosx

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Higher-Order Derivatives of 9sinx

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 9sin(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 9sin(x), we generally use f⁽ⁿ⁾(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:

When x is 0, the derivative of 9sinx = 9cos(0), which is 9. The function 9sinx is continuous and differentiable for all real x.

Common Mistakes and How to Avoid Them in Derivatives of 9sinx

Students frequently make mistakes when differentiating 9sinx. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:

Problem 1

Calculate the derivative of (9sinx · cosx)

Okay, lets begin

Here, we have f(x) = 9sinx · cosx. Using the product rule, f'(x) = u′v + uv′ In the given equation, u = 9sinx and v = cosx. Let’s differentiate each term, u′ = d/dx (9sinx) = 9cosx v′ = d/dx (cosx) = -sinx Substituting into the given equation, f'(x) = (9cosx) · (cosx) + (9sinx) · (-sinx) Let’s simplify terms to get the final answer, f'(x) = 9cos²x - 9sin²x Thus, the derivative of the specified function is 9(cos²x - sin²x).

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.

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Problem 2

A new amusement park ride moves in a wave-like motion. The height of the ride at a given time t is represented by h(t) = 9sin(t). If t = π/6 seconds, determine the rate of change of the height of the ride.

Okay, lets begin

We have h(t) = 9sin(t) (height of the ride)...(1) Now, we will differentiate the equation (1) Take the derivative 9sin(t): dh/dt = 9cos(t) Given t = π/6, substitute this into the derivative: dh/dt = 9cos(π/6) dh/dt = 9 * (√3/2) dh/dt = 9√3/2 Hence, the rate of change of the height of the ride at t = π/6 seconds is 9√3/2.

Explanation

We find the rate of change of the height of the ride at t= π/6 as 9√3/2, which means that at this time, the height is increasing at this rate.

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Problem 3

Derive the second derivative of the function h(t) = 9sin(t).

Okay, lets begin

The first step is to find the first derivative, dh/dt = 9cos(t)...(1) Now we will differentiate equation (1) to get the second derivative: d²h/dt² = d/dt [9cos(t)] d²h/dt² = -9sin(t) Therefore, the second derivative of the function h(t) = 9sin(t) is -9sin(t).

Explanation

We use the step-by-step process, where we start with the first derivative. We then differentiate cos(t) to get the second derivative.

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Problem 4

Prove: d/dt (9sin²(t)) = 18sin(t)cos(t).

Okay, lets begin

Let’s start using the chain rule: Consider y = 9sin²(t) = 9(sin(t))² To differentiate, we use the chain rule: dy/dt = 9 * 2sin(t) * d/dt [sin(t)] Since the derivative of sin(t) is cos(t), dy/dt = 18sin(t)cos(t) Substituting y = 9sin²(t), d/dt (9sin²(t)) = 18sin(t)cos(t) Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace sin(t) with its derivative. As a final step, we substitute y = 9sin²(t) to derive the equation.

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Problem 5

Solve: d/dx (9sinx/x)

Okay, lets begin

To differentiate the function, we use the quotient rule: d/dx (9sinx/x) = (d/dx (9sinx) · x - 9sinx · d/dx(x))/x² We will substitute d/dx (9sinx) = 9cosx and d/dx (x) = 1 = (9cosx · x - 9sinx · 1) / x² = (9xcosx - 9sinx) / x² Therefore, d/dx (9sinx/x) = (9xcosx - 9sinx) / x²

Explanation

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.

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FAQs on the Derivative of 9sinx

1.Find the derivative of 9sinx.

Using the derivative of sinx, we have: d/dx (9sinx) = 9cosx

2.Can we use the derivative of 9sinx in real life?

Yes, we can use the derivative of 9sinx in real life to analyze oscillatory motion, which is common in physics and engineering problems.

3.Is it possible to take the derivative of 9sinx at any point?

Yes, 9sinx is differentiable at all points on the real number line, so it is possible to take the derivative at any point.

4.What rule is used to differentiate 9sinx?

We use the constant multiple rule and the standard derivative of sinx to differentiate 9sinx: d/dx (9sinx) = 9cosx.

5.Are the derivatives of 9sinx and sinx the same?

No, they are not the same. The derivative of sinx is cosx, while the derivative of 9sinx is 9cosx due to the constant multiple.

Important Glossaries for the Derivative of 9sinx

Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x. Sine Function: The sine function is a primary trigonometric function and is written as sinx. Constant Multiple Rule: A rule in differentiation that states the derivative of a constant times a function is the constant times the derivative of the function. Cosine Function: A trigonometric function related to the sine function, written as cosx. Chain Rule: A rule for differentiating compositions of functions, used extensively in calculus.

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Jaskaran Singh Saluja

About the Author

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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