Derivative of sin(xy)
2026-02-28 06:20 Diff
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Last updated on September 22, 2025

We explore the derivative of sin(xy) using implicit differentiation, which is essential for understanding how the function changes as both x and y vary. This concept is particularly useful in multivariable calculus and real-life applications involving related rates. We will now discuss the derivative of sin(xy) in detail.

What is the Derivative of sin(xy)?

To understand the derivative of sin(xy), we need to use implicit differentiation because sin(xy) is a composite function of two variables. It is commonly represented as d/dx (sin(xy)) or (sin(xy))'. The function sin(xy) has a derivative that depends on both x and y, illustrating its differentiability within its domain. The key concepts involve: -

Sine Function: sin(xy) is a trigonometric function involving the product of x and y. -

Product Rule: Necessary for differentiating xy since it's a product of two variables. 

Chain Rule: Used to differentiate composite functions like sin(xy).

Derivative of sin(xy) Formula

The derivative of sin(xy) with respect to x can be expressed using implicit differentiation.

The formula is: d/dx (sin(xy)) = cos(xy) * (y + x (dy/dx))

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

Proofs of the Derivative of sin(xy)

We can derive the derivative of sin(xy) using implicit differentiation. To show this, we will use trigonometric identities along with differentiation rules. Here are some methods to prove it:

Using Implicit Differentiation

To find the derivative of sin(xy), we apply implicit differentiation. Assume z = sin(xy).

Differentiate both sides with respect to x: d/dx (z) = d/dx (sin(xy)) dz/dx = cos(xy) * (d/dx (xy))

Using the product rule on xy: d/dx (xy) = y + x (dy/dx) Thus, dz/dx = cos(xy) * (y + x (dy/dx))

Using Chain Rule

To differentiate sin(xy) using the chain rule, consider u = xy and v = sin(u).

Differentiate v with respect to x: dv/dx = cos(u) * du/dx Since u = xy, apply the product rule: du/dx = y + x (dy/dx)

Thus, dv/dx = cos(xy) * (y + x (dy/dx))

Using Product Rule

To differentiate sin(xy) using the product rule, express it as a product: sin(xy) = sin(u), where u = xy.

Apply product and chain rules: d/dx (sin(u)) = cos(u) * du/dx du/dx = y + x (dy/dx)

Thus, d/dx (sin(xy)) = cos(xy) * (y + x (dy/dx))

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Higher-Order Derivatives of sin(xy)

Higher-order derivatives involve differentiating a function multiple times. For functions like sin(xy), understanding higher-order derivatives can be complex but insightful.

For instance, if sin(xy) describes a wave, the first derivative gives the slope, and the second derivative indicates concavity or convexity.

The nth derivative, denoted as fⁿ(x), provides information about the rate of change at higher levels.

Special Cases:

When x or y is such that xy = π/2, the derivative includes cos(π/2), which results in 0, simplifying the expression. When x = 0 or y = 0, the derivative simplifies because sin(0) = 0.

Common Mistakes and How to Avoid Them in Derivatives of sin(xy)

Students often make mistakes when differentiating sin(xy). These errors can be rectified by understanding the correct methods. Here are some common mistakes and solutions:

Problem 1

Calculate the derivative of sin(3xy).

Okay, lets begin

Let z = sin(3xy). Using implicit differentiation, dz/dx = cos(3xy) * d/dx (3xy)

Apply the product rule: d/dx (3xy) = 3(y + 3x(dy/dx))

Substitute back: dz/dx = cos(3xy) * 3(y + 3x(dy/dx))

Thus, the derivative of sin(3xy) is 3 cos(3xy) * (y + 3x(dy/dx)).

Explanation

We find the derivative by differentiating sin(3xy) using the chain rule and product rule. This involves differentiating the inner function 3xy and applying the chain rule.

Well explained 👍

Problem 2

A circular track is expanding, and its radius is given by the function r = sin(xy). If x = 2 meters and y = 1 meter, find dr/dx.

Okay, lets begin

Given r = sin(xy), differentiate with respect to x: dr/dx = cos(xy) * d/dx (xy)

Using the product rule: d/dx (xy) = y + x(dy/dx)

Substitute x = 2 and y = 1: dr/dx = cos(2 * 1) * (1 + 2(dy/dx))

Since dy/dx is not specified, assume dy/dx = 0 for simplicity: dr/dx = cos(2) * 1

Thus, dr/dx = cos(2).

Explanation

We find dr/dx by differentiating the given function implicitly, substituting given values, and assuming dy/dx = 0 for simplicity.

Well explained 👍

Problem 3

Derive the second derivative of the function z = sin(xy).

Okay, lets begin

First, find the first derivative: dz/dx = cos(xy) * (y + x(dy/dx))... (1)

Now differentiate again: d²z/dx² = d/dx [cos(xy) * (y + x(dy/dx))]

Apply product and chain rules to (1): = -sin(xy) * (y + x(dy/dx)) * (y + x(dy/dx)) + cos(xy) * (dy/dx + x(d²y/dx²))

Simplify terms: d²z/dx² = -sin(xy) * (y + x(dy/dx))² + cos(xy) * (dy/dx + x(d²y/dx²))

Explanation

To derive the second derivative, we apply implicit differentiation to the first derivative, using the product and chain rules to simplify.

Well explained 👍

Problem 4

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

Okay, lets begin

Consider w = sin²(xy). Using the chain rule: dw/dx = 2 sin(xy) * d/dx (sin(xy))

Differentiate sin(xy): d/dx (sin(xy)) = cos(xy) * (y + x(dy/dx))

Substitute: dw/dx = 2 sin(xy) cos(xy) * (y + x(dy/dx)) Hence proved.

Explanation

In this process, we apply the chain rule to differentiate sin²(xy) and substitute the derivative of sin(xy) to complete the proof.

Well explained 👍

Problem 5

Solve: d/dx (sin(xy)/x).

Okay, lets begin

To differentiate the function, use the quotient rule: d/dx (sin(xy)/x) = (d/dx (sin(xy)) * x - sin(xy) * d/dx(x))/x² Differentiate sin(xy): d/dx (sin(xy)) = cos(xy) * (y + x(dy/dx))

Substitute: = (cos(xy) * (y + x(dy/dx)) * x - sin(xy))/x² Simplify: = x cos(xy) * (y + x(dy/dx)) - sin(xy)/x²

Thus, d/dx (sin(xy)/x) = [x cos(xy) * (y + x(dy/dx)) - sin(xy)]/x²

Explanation

We differentiate the given function using the quotient rule, simplifying to obtain the final result.

Well explained 👍

FAQs on the Derivative of sin(xy)

1.Find the derivative of sin(xy).

Using implicit differentiation on sin(xy), we apply the chain rule: d/dx (sin(xy)) = cos(xy) * (y + x(dy/dx))

2.Can the derivative of sin(xy) be used in real life?

Yes, derivatives of functions like sin(xy) can be applied in real-life scenarios involving rates of change in physics, engineering, and other fields.

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

Yes, at xy = π/2, the cosine term in the derivative becomes 0, greatly simplifying the expression.

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

We use the quotient rule to differentiate sin(xy)/x, resulting in: d/dx (sin(xy)/x) = [x cos(xy) * (y + x(dy/dx)) - sin(xy)]/x²

5.Are the derivatives of sin(xy) and sin(yx) the same?

Important Glossaries for the Derivative of sin(xy)

  • Derivative: The derivative of a function describes how the function changes in response to changes in the variables.
  • Implicit Differentiation: A technique used to find derivatives of functions that are not explicitly solved for one variable in terms of another.
  • Product Rule: A rule used to differentiate products of two functions.
  • Chain Rule: A rule used for differentiating composite functions.
  • Trigonometric Functions: Functions like sine and cosine that are fundamental in calculus and analysis.

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Jaskaran Singh Saluja

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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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