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Last updated on October 17, 2025
The derivative of cos(y), which is -sin(y), is a fundamental concept in calculus. It helps us understand how the cosine function changes in response to a slight change in y. Derivatives are essential in various real-life applications, like physics and engineering. We will now explore the derivative of cos(y) in detail.
The derivative of cos y is commonly represented as d/dy (cos y) or (cos y)', and its value is -sin(y). This indicates that the cosine function is differentiable within its domain.
Key concepts include:
Cosine Function: cos(y) is one of the fundamental trigonometric functions.
Derivative Rule: Rule for differentiating cos(y) which results in -sin(y).
Sine Function: sin(y) is another primary trigonometric function, related to cos(y).
The derivative of cos y can be denoted as d/dy (cos y) or (cos y)'.
The formula we use to differentiate cos y is: d/dy (cos y) = -sin(y) (cos y)' = -sin(y)
The formula applies to all y where the cosine function is defined.
We can derive the derivative of cos y using proofs. To show this, we will use trigonometric identities and the rules of differentiation.
Here are several methods we use to prove this:
We will now demonstrate that the differentiation of cos y results in -sin(y) using these methods:
The derivative of cos y can be proved using the First Principle, expressing the derivative as the limit of the difference quotient. Consider f(y) = cos y. Its derivative can be expressed as: f'(y) = limₕ→₀ [f(y + h) - f(y)] / h Given f(y) = cos y, we write f(y + h) = cos(y + h). Substituting these into the equation: f'(y) = limₕ→₀ [cos(y + h) - cos y] / h = limₕ→₀ [-sin(y + h/2) sin(h/2)] / [h] = -limₕ→₀ [sin(h/2) / (h/2) sin(y + h/2)] Using limit formulas, limₕ→₀ [sin(h/2) / (h/2)] = 1. f'(y) = -sin(y) Hence, proved.
To prove the differentiation of cos y using the chain rule, we use the identity: cos y = cos(y) By the chain rule: d/dy (cos y) = -sin(y)
We will now prove the derivative of cos y using the product rule. Let u = 1 and v = cos y, then d/dy (cos y) = u'v + uv'. u' = 0 and v' = -sin(y). Substituting these into the product rule gives: d/dy (cos y) = 0·cos y + 1·(-sin(y)) = -sin(y). Thus, the derivative of cos y is -sin(y).
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives provide deeper insights into the behavior of functions like cos(y). For the first derivative, we write f′(y), indicating how the function changes or its slope at a certain point.
The second derivative, f′′(y), is derived from the first derivative. Similarly, the third derivative, f′′′(y), is the result of the second derivative, and this pattern continues.
For the nth Derivative of cos(y), we use fⁿ(y) for the nth derivative, indicating the change in the rate of change.
When y is π/2, the derivative is -sin(π/2), which is -1.
When y is 0, the derivative of cos y is -sin(0), which is 0.
Students frequently make mistakes when differentiating cos y. Understanding the correct solutions can help resolve these mistakes. Here are a few common mistakes and ways to solve them:
Calculate the derivative of cos(y)·sin(y)
Here, we have f(y) = cos(y)·sin(y). Using the product rule, f'(y) = u′v + uv′ In the given equation, u = cos(y) and v = sin(y). Differentiate each term: u′ = d/dy (cos(y)) = -sin(y) v′ = d/dy (sin(y)) = cos(y) Substituting into the given equation, f'(y) = (-sin(y))·sin(y) + cos(y)·cos(y) = -sin²(y) + cos²(y) Thus, the derivative of cos(y)·sin(y) is cos²(y) - sin²(y).
We find the derivative of the given function by dividing it into two parts.
First, we find its derivative, then combine them using the product rule to get the final result.
The temperature T in a room changes according to the function T = cos(y), where y is the angle of the sun in degrees. If y = 60°, calculate the rate of change of temperature.
We have T = cos(y) (temperature change)...(1) Now, we will differentiate the equation (1) Take the derivative of cos(y): dT/dy = -sin(y) Given y = 60° dT/dy = -sin(60°) = -√3/2 Hence, the rate of change of temperature when y = 60° is -√3/2.
We find the rate of temperature change at y = 60° using the derivative -sin(y).
This indicates that the temperature decreases at a rate of -√3/2 at this angle.
Derive the second derivative of the function T = cos(y).
The first step is to find the first derivative, dT/dy = -sin(y)...(1) Now, differentiate equation (1) to get the second derivative: d²T/dy² = d/dy [-sin(y)] = -cos(y) Therefore, the second derivative of the function T = cos(y) is -cos(y).
We use a step-by-step process, starting with the first derivative.
We then differentiate -sin(y) to find the second derivative, which is -cos(y).
Prove: d/dy (cos²(y)) = -2 cos(y) sin(y).
Let's start using the chain rule: Consider T = cos²(y) = [cos(y)]² To differentiate, use the chain rule: dT/dy = 2 cos(y)·d/dy [cos(y)] Since the derivative of cos(y) is -sin(y), dT/dy = 2 cos(y)·(-sin(y)) = -2 cos(y) sin(y) Hence proved.
In this step-by-step process, we use the chain rule to differentiate the equation.
We replace cos(y) with its derivative, -sin(y), and simplify to derive the equation.
Solve: d/dy (cos y/y)
To differentiate the function, use the quotient rule: d/dy (cos y/y) = [d/dy (cos y)·y - cos y·d/dy(y)] / y² Substitute d/dy (cos y) = -sin(y) and d/dy (y) = 1 = [-sin(y)·y - cos(y)·1] / y² = [-y sin(y) - cos(y)] / y² Therefore, d/dy (cos y/y) = [-y sin(y) - cos(y)] / y²
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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