Last updated on September 6, 2025
We use the derivative of cos(x), which is -sin(x), as a measuring tool for how the cosine function changes in response to a slight change in x. Derivatives help us calculate various rates of change in real-life situations. We will now talk about the derivative of cos(x) in detail.
We now understand the derivative of cos x.
It is commonly represented as d/dx (cos x) or (cos x)', and its value is -sin x.
The function cos x has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Cosine Function: cos(x) is one of the primary trigonometric functions.
Derivative Rule: Rule for differentiating cos(x).
Sine Function: sin(x) is another primary trigonometric function.
The derivative of cos x can be denoted as d/dx (cos x) or (cos x)'. The formula we use to differentiate cos x is: d/dx (cos x) = -sin x (or) (cos x)' = -sin x The formula applies to all x.
We can derive the derivative of cos x 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
Using Product Rule
We will now demonstrate that the differentiation of cos x results in -sin x using the above-mentioned methods:
By First Principle
The derivative of cos x can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient.
To find the derivative of cos x using the first principle, we will consider f(x) = cos x.
Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)
Given that f(x) = cos x, we write f(x + h) = cos (x + h).
Substituting these into equation (1), f'(x) = limₕ→₀ [cos(x + h) - cos x] / h = limₕ→₀ [ [-sin(x + h) sin x - cos(x + h) cos x] / h]
We now use the formula for cosine subtraction: cos A - cos B = -2sin((A + B)/2)sin((A - B)/2). f'(x) = limₕ→₀ [ -2sin((x + h + x)/2)sin((x + h - x)/2) ] / h = limₕ→₀ [ -2sin(x + h/2)sin(h/2) ] / h = limₕ→₀ -sin(x) [ sin(h/2) / (h/2) ] · 1/2
Using limit formulas, limₕ→₀ (sin(h/2))/(h/2) = 1. f'(x) = -sin(x) · 1 · 1 f'(x) = -sin x
Hence, proved.
Using Chain Rule
To prove the differentiation of cos x using the chain rule,
We use the formula: cos x = 1/sin x Consider f(x) = 1 and g(x) = sin x
So we get, cos x = f(x)/g(x) By quotient rule: d/dx [f(x) / g(x)] = [f'(x) g(x) - f(x) g'(x)] / [g(x)]²… (1)
Let’s substitute f(x) = 1 and g(x) = sin x in equation (1), d/dx (cos x) = [(0) (sin x) - (1) (cos x)] / (sin x)² = -cos x / sin² x …(2)
Here, we use the identity sin² x + cos² x = 1.
Substituting this into (2), d/dx (cos x) = -sin x. Since sin x = sin x, we write: d/dx(cos x) = -sin x
Using Product Rule
We will now prove the derivative of cos x using the product rule.
The step-by-step process is demonstrated below:
Here, we use the formula, cos x = 1/sin x cos x = (1) · (sin x)⁻¹
Given that, u = 1 and v = (sin x)⁻¹
Using the product rule formula: d/dx [u.v] = u'. v + u. v' u' = d/dx (1) = 0 (substitute u = 1)
Here we use the chain rule: v = (sin x)⁻¹ = (sin x)⁻¹ (substitute v = (sin x)⁻¹) v' = -1 · (sin)⁻² · d/dx (sin x) v' = -cos x / sin² x Again, use the product rule formula: d/dx (cos x) = u'. v + u. v'
Let’s substitute u = 1, u' = 0, v = (sin x)⁻¹, and v' = -cos x / sin² x
When we simplify each term: We get, d/dx (cos x) = -sin x
Thus: d/dx (cos x) = -sin x.
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).
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), 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 the x is π/2, the derivative is -sin(π/2), which is -1. When the x is 0, the derivative of cos x = -sin(0), which is 0.
Students frequently make mistakes when differentiating cos x. 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·sin x)
Here, we have f(x) = cos x·sin x.
Using the product rule, f'(x) = u′v + uv′
In the given equation, u = cos x and v = sin x.
Let’s differentiate each term, u′= d/dx (cos x) = -sin x v′= d/dx (sin x) = cos x substituting into the given equation, f'(x) = (-sin x)·(sin x) + (cos x)·(cos x)
Let’s simplify terms to get the final answer, f'(x) = -sin²x + cos²x
Thus, the derivative of the specified function is -sin²x + cos²x.
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.
XYZ Telecom sponsored the construction of a bridge. The elevation is represented by the function y = cos(x) where y represents the height of the bridge at a distance x. If x = π/3 meters, measure the slope of the bridge.
We have y = cos (x) (slope of the bridge)...(1)
Now, we will differentiate the equation (1)
Take the derivative cos(x): dy/dx = -sin(x)
Given x = π/3 (substitute this into the derivative) -sin(π/3) = -√3/2
Hence, we get the slope of the bridge at a distance x = π/3 as -√3/2.
We find the slope of the bridge at x = π/3 as -√3/2, which means that at a given point, the height of the bridge would decrease at a rate √3/2 times the horizontal distance.
Derive the second derivative of the function y = cos(x).
The first step is to find the first derivative, dy/dx = -sin(x)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [-sin(x)]
Here we use the derivative of sine, d²y/dx² = -cos(x)
Therefore, the second derivative of the function y = cos(x) is -cos(x).
We use the step-by-step process, where we start with the first derivative.
Using the derivative of sine, we find the second derivative.
We then substitute the identity and simplify the terms to find the final answer.
Prove: d/dx (cos²(x)) = -2 cos(x) sin(x).
Let’s start using the chain rule:
Consider y = cos²(x) [cos(x)]²
To differentiate, we use the chain rule: dy/dx = 2 cos(x) · d/dx [cos(x)]
Since the derivative of cos(x) is -sin(x), dy/dx = 2 cos(x) · (-sin(x))
Substituting y = cos²(x), d/dx (cos²(x)) = -2 cos(x) · sin(x) Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace cos(x) with its derivative. As a final step, we substitute y = cos²(x) to derive the equation.
Solve: d/dx (cos x/x)
To differentiate the function, we use the quotient rule:
d/dx (cos x/x) = (d/dx (cos x) · x - cos x · d/dx(x))/ x²
We will substitute d/dx (cos x) = -sin x and d/dx (x) = 1 (-sin x · x - cos x)/ x² = (-x sin x - cos x)/ x²
Therefore, d/dx (cos x/x) = (-x sin x - cos x)/ 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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