Derivative of sin(x+y)
2026-02-28 01:25 Diff
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Last updated on October 17, 2025

We use the derivative of sin(x+y) to measure how the sine function changes in response to a slight change in the variables x and y. Derivatives are vital in calculating various changes in real-life scenarios. We will now discuss the derivative of sin(x+y) in detail.

What is the Derivative of sin(x+y)?

The derivative of sin(x+y) is commonly represented as d/dx (sin(x+y)) or (sin(x+y))'. Its value is cos(x+y), indicating it is differentiable within its domain.

The key concepts are mentioned below: 

Sine Function: sin(x+y) represents the sine of the sum of two angles. 

Chain Rule: A crucial rule for differentiating composite functions like sin(x+y). 

Cosine Function: cos(x+y) is part of the derivative of sin(x+y).

Derivative of sin(x+y) Formula

The derivative of sin(x+y) can be denoted as d/dx (sin(x+y)) or (sin(x+y))'.

The formula we use to differentiate sin(x+y) is: d/dx (sin(x+y)) = cos(x+y) * (dx/dx + dy/dx)

The formula applies to all x and y where the function is defined.

Proofs of the Derivative of sin(x+y)

We can derive the derivative of sin(x+y) using various proofs. To show this, we will use trigonometric identities along with the rules of differentiation.

There are several methods we use to prove this, such as:

By First Principle

The derivative of sin(x+y) can be proved using the First Principle, which expresses the derivative as the limit of the difference quotient. To find the derivative of sin(x+y) using the first principle, consider f(x) = sin(x+y). Its derivative can be expressed as the following limit: f'(x) = limₕ→₀ [f(x + h) - f(x)] / h Given that f(x) = sin(x+y), we write f(x + h) = sin(x+h+y). Substituting these into the equation, f'(x) = limₕ→₀ [sin(x+h+y) - sin(x+y)] / h Using the identity sin(A) - sin(B) = 2 * cos((A+B)/2) * sin((A-B)/2), f'(x) = limₕ→₀ [2 * cos((2x+h+2y)/2) * sin(h/2)] / h = limₕ→₀ [cos(x+h/2+y) * sin(h/2)] / (h/2) Using limit formulas, limₕ→₀ (sin(h/2))/(h/2) = 1. f'(x) = cos(x+y) Hence, proved.

Using Chain Rule

To prove the differentiation of sin(x+y) using the chain rule, Let u = x+y, then sin(u) = sin(x+y). The derivative using the chain rule is: d/du (sin u) * du/dx = cos(u) * (d(x+y)/dx) = cos(x+y) * (1 + dy/dx) Using Product Rule We can also use the product rule to prove the derivative of sin(x+y) by considering a function of two variables. The step-by-step process is: Let u = x and v = y, then sin(x+y) = sin(u+v). Using the product rule: d/dx [sin(u+v)] = d/du [sin(u+v)] * du/dx + d/dv [sin(u+v)] * dv/dx = cos(u+v) * (1 + dy/dx) = cos(x+y) * (1 + dy/dx) Thus, the derivative is derived.

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Higher-Order Derivatives of sin(x+y)

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+y).

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+y), 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

When x+y = π/2, the derivative is zero because cos(π/2) = 0.

When x+y = 0, the derivative of sin(x+y) = cos(0), which is 1.

Common Mistakes and How to Avoid Them in Derivatives of sin(x+y)

Students frequently make mistakes when differentiating sin(x+y). 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 (sin(x+y)·cos(x+y))

Okay, lets begin

Here, we have f(x) = sin(x+y)·cos(x+y). Using the product rule, f'(x) = u′v + uv′ In the given equation, u = sin(x+y) and v = cos(x+y). Let’s differentiate each term, u′ = d/dx (sin(x+y)) = cos(x+y) * (1 + dy/dx) v′ = d/dx (cos(x+y)) = -sin(x+y) * (1 + dy/dx) Substituting into the given equation, f'(x) = (cos(x+y) * (1 + dy/dx)). (cos(x+y)) + (sin(x+y)). (-sin(x+y) * (1 + dy/dx)) Let’s simplify terms to get the final answer, f'(x) = cos²(x+y) * (1 + dy/dx) - sin²(x+y) * (1 + dy/dx) Thus, the derivative of the specified function is cos²(x+y) * (1 + dy/dx) - sin²(x+y) * (1 + dy/dx).

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.

Well explained 👍

Problem 2

A company wants to model the combined effect of two factors on sales using the function s = sin(x+y), where s represents sales, x is advertising, and y is market conditions. If x+y = π/4, find how sales change with respect to x.

Okay, lets begin

We have s = sin(x+y) (sales model)...(1) Now, we will differentiate equation (1) Take the derivative sin(x+y): ds/dx = cos(x+y) * (1 + dy/dx) Given x+y = π/4 (substitute this into the derivative) cos(π/4) = 1/√2 Hence, ds/dx = 1/√2 * (1 + dy/dx) The change in sales with respect to x at x+y = π/4 is 1/√2 * (1 + dy/dx).

Explanation

We find the change in sales with respect to advertising and market conditions at x+y = π/4 using the derivative.

This shows how sensitive sales are to changes in these factors.

Well explained 👍

Problem 3

Derive the second derivative of the function y = sin(x+y).

Okay, lets begin

The first step is to find the first derivative, dy/dx = cos(x+y) * (1 + dy/dx)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [cos(x+y) * (1 + dy/dx)] Here we use the product rule, d²y/dx² = -sin(x+y) * (1 + dy/dx) * (1 + dy/dx) + cos(x+y) * d²y/dx² Therefore, the second derivative of the function y = sin(x+y) is -sin(x+y) * (1 + dy/dx)² + cos(x+y) * d²y/dx².

Explanation

We use a step-by-step process, starting with the first derivative.

Using the product rule, we differentiate cos(x+y).

We then substitute the identity and simplify the terms to find the final answer.

Well explained 👍

Problem 4

Prove: d/dx (sin²(x+y)) = 2 sin(x+y) cos(x+y) * (1 + dy/dx).

Okay, lets begin

Let’s start using the chain rule: Consider y = sin²(x+y) = [sin(x+y)]² To differentiate, we use the chain rule: dy/dx = 2 sin(x+y) * d/dx [sin(x+y)] Since the derivative of sin(x+y) is cos(x+y) * (1 + dy/dx), dy/dx = 2 sin(x+y) * cos(x+y) * (1 + dy/dx) Substituting y = sin²(x+y), d/dx (sin²(x+y)) = 2 sin(x+y) cos(x+y) * (1 + dy/dx) Hence proved.

Explanation

In this step-by-step process, we used the chain rule to differentiate the equation.

Then, we replace sin(x+y) with its derivative.

As a final step, we substitute y = sin²(x+y) to derive the equation.

Well explained 👍

Problem 5

Solve: d/dx (sin(x+y)/x)

Okay, lets begin

To differentiate the function, we use the quotient rule: d/dx (sin(x+y)/x) = (d/dx (sin(x+y)) * x - sin(x+y) * d/dx(x))/x² We will substitute d/dx (sin(x+y)) = cos(x+y) * (1 + dy/dx) and d/dx (x) = 1 = (cos(x+y) * (1 + dy/dx) * x - sin(x+y) * 1) / x² = (x * cos(x+y) * (1 + dy/dx) - sin(x+y)) / x² Therefore, d/dx (sin(x+y)/x) = (x * cos(x+y) * (1 + dy/dx) - sin(x+y)) / 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.

Well explained 👍

FAQs on the Derivative of sin(x+y)

1.Find the derivative of sin(x+y).

Using the chain rule on sin(x+y), d/dx (sin(x+y)) = cos(x+y) * (1 + dy/dx).

2.Can we use the derivative of sin(x+y) in real life?

Yes, we can use the derivative of sin(x+y) in real life to calculate changes in various fields such as physics and engineering, especially when analyzing wave interference patterns.

3.Is it possible to take the derivative of sin(x+y) at the point where x+y = π/2?

Yes, at x+y = π/2, cos(x+y) = 0, so the derivative of sin(x+y) is 0.

4.What rule is used to differentiate sin(x+y)/x?

We use the quotient rule to differentiate sin(x+y)/x, d/dx (sin(x+y)/x) = (x * cos(x+y) * (1 + dy/dx) - sin(x+y)) / x².

5.Are the derivatives of sin(x+y) and sin⁻¹(x) the same?

No, they are different. The derivative of sin(x+y) is cos(x+y) * (1 + dy/dx), while the derivative of sin⁻¹(x) is 1/√(1-x²).

Important Glossaries for the Derivative of sin(x+y)

  • Derivative: The derivative of a function indicates how the given function changes in response to a slight change in x.
  • Sine Function: A primary trigonometric function, often denoted as sin(x).
  • Cosine Function: A trigonometric function, the derivative of sine, denoted as cos(x).
  • Chain Rule: A rule for finding the derivative of composite functions.
  • Quotient Rule: A method used to differentiate functions that are ratios of two differentiable functions.

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