Question

Find \(d y / d x\) by implicit differentiation. \(y=\sin x y\)

Short answer

The derivative \(dy/dx\) is \(y \cdot \cos(xy) / (1 - x \cdot \cos(xy))\).

Step-by-step solution

- Rewrite the equation

Allow for differentiation by rewriting the equation as \(y-\sin(xy)=0\).

- Apply the derivative

Take the derivative of both sides with respect to \(x\). Use the chain rule, remembering that \(y\) is also a function of \(x\). Applying these rules, we obtain \(dy/dx - (\cos(xy) \cdot (x \cdot dy/dx + y)) = 0\).

- Isolate dy/dx

Rearrange the equation to isolate \(dy/dx\). This will result in \(dy/dx \cdot (1 - x \cdot \cos(xy))= y \cdot \cos(xy)\). Divide both sides by \(1 - x \cdot \cos(xy)\) to solve for \(dy/dx\).

- Write out the solution

After simplifying, the derivative of \(y(x)\) with respect to \(x\), \(dy/dx\), is given by \(dy/dx = y*cos(xy)/(1 - x \cdot \cos(xy))\).

Key concepts

Chain Rule

The chain rule is a fundamental principle in calculus used for finding the derivative of a composite function. Imagine you have a function f that is dependent on g, and g is in turn dependent on x. The chain rule allows us to calculate the derivative of f with respect to x by multiplying the derivative of f with respect to g by the derivative of g with respect to x. The symbolic representation is \( \frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \).

When applied to our exercise involving trigonometric functions and implicit differentiation, the chain rule is critically important. As seen in the problem, the derivative of \( y = \sin(x y) \) with respect to x involves treating y as a function of x. This means when we take the derivative of \( \sin(x y) \), we first differentiate \( \sin \) with respect to its argument (the inner function \( x y \) in this case), and then multiply it by the derivative of the inner function (\( x y \) here) with respect to x.

Understanding the chain rule is essential, not only for calculus but for all advanced math, as it is widely used across various problems involving composite functions.

Derivative of Trigonometric Functions

The derivative of standard trigonometric functions is a topic covered early in calculus. It’s vital to memorize these derivatives as they frequently arise in various types of problems, including those with implicit differentiation. For instance, the derivative of \( \sin(x) \) with respect to \( x \) is \( \cos(x) \), and similarly, the derivative of \( \cos(x) \) is \( -\sin(x) \).

In the context of our problem, where we need to find the derivative of \( y = \sin(x y) \) implicitly, it's important to apply these standard derivatives correctly. Since \( y \) is a function of \( x \) as suggested by the presence of \( dy/dx \) in our solution, every time you differentiate a term involving \( y \) in the function, you must also include \( dy/dx \) as part of your resulting expression. This ensures that the derivatives are calculated fully, taking into account the inherent relationship between \( y \) and \( x \) in the equation.

Isolating Derivatives

The process of isolating the derivative in an implicit differentiation problem is akin to solving a puzzle where you need to find the piece marked \( dy/dx \) and make it the subject of the equation. This often requires algebraic manipulations such as addition, subtraction, multiplication, division, or factoring.

In the given exercise, once we apply the derivative to both sides of the equation using the chain rule, we end up with terms involving \( dy/dx \) that are mixed with other terms. To find \( dy/dx \) explicitly, we need to collect all terms involving \( dy/dx \) on one side of the equals sign and move everything else to the other side. Here, we isolate \( dy/dx \) by factoring it out and then dividing both sides by the remaining expression.

This can be seen in step 3 of our solution, where after rearranging, we divide through by \( 1 - x \cdot \cos(xy) \) to get \( dy/dx \). Understanding how to isolate \( dy/dx \) is crucial for solving implicit differentiation problems, and it requires both a good grasp of algebra and the ability to recognize how to manipulate the terms involving derivatives.