Differentiate cos(xy)

Understand the Problem

The question is asking for the differentiation of the function cos(xy) with respect to the variable x. This involves applying the chain rule and product rule of differentiation as it is a function of two variables, x and y.

Answer

$$ \frac{dz}{dx} = -\sin(xy) \left( y + x \frac{dy}{dx} \right) $$

Steps to Solve

Identify the function We have the function $z = \cos(xy)$, which depends on both $x$ and $y$.
Use the chain rule We differentiate $z$ with respect to $x$. According to the chain rule, we need to differentiate the outer function and multiply it by the derivative of the inner function.

The derivative of $\cos(u)$ where $u = xy$ is: $$ \frac{dz}{du} = -\sin(u) = -\sin(xy) $$
Differentiate the inner function using the product rule Now, we need to find $\frac{du}{dx}$ where $u = xy$. Here, we apply the product rule which states: $$ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} $$

In our case, $u = x$ and $v = y$, so: $$ \frac{du}{dx} = y \cdot \frac{d(x)}{dx} + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx} $$
Combine results Now we can combine the derivatives using the chain rule: $$ \frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx} $$ Replacing the derivatives we found: $$ \frac{dz}{dx} = -\sin(xy) \cdot \left( y + x \frac{dy}{dx} \right) $$

The derivative of $z = \cos(xy)$ with respect to $x$ is: $$ \frac{dz}{dx} = -\sin(xy) \left( y + x \frac{dy}{dx} \right) $$

More Information

This result illustrates the application of both the chain rule and product rule in multivariable calculus. Differentiating functions that contain multiple variables is a common scenario in calculus, demonstrating the interconnectedness of variables.

Tips

Forgetting to apply the product rule when differentiating a function that is a product of two variables.
Not considering that $y$ may be a function of $x$, leading to incorrect assumptions about its derivative.

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