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128 LearnersLast 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.
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).
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.
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:
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))
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))
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))
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.
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.
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:
Calculate the derivative of sin(3xy).
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)).
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.
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.
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).
We find dr/dx by differentiating the given function implicitly, substituting given values, and assuming dy/dx = 0 for simplicity.
Derive the second derivative of the function z = sin(xy).
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))
To derive the second derivative, we apply implicit differentiation to the first derivative, using the product and chain rules to simplify.
Prove: d/dx (sin²(xy)) = 2 sin(xy) cos(xy) * (y + x(dy/dx)).
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.
In this process, we apply the chain rule to differentiate sin²(xy) and substitute the derivative of sin(xy) to complete the proof.
Solve: d/dx (sin(xy)/x).
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
We differentiate the given function using the quotient rule, simplifying 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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