Question

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$\sin x y=x+y$$

Short answer

Question: Find the derivative of the implicitly-defined function \(\sin(xy) = x + y\). Answer: The derivative of the implicitly-defined function is \(\frac{dy}{dx} = \frac{1 - yx'}{yx+1}\).

Step-by-step solution

Differentiate both sides of the equation with respect to x

Start by differentiating both sides of the equation $$\sin(xy)=x+y$$ with respect to x. Remember, whenever you differentiate a term with y, apply the chain rule and multiply it by \(\frac{dy}{dx}\). Differentiating both sides: $$\frac{d(\sin(xy))}{dx} = \frac{d(x+y)}{dx}$$.

Apply the chain rule and product rule where needed

Now, we need to apply the chain rule and product rule where needed, find the derivative of each term: Differentiating \(\sin(xy)\) with respect to x using chain rule and product rule: $$\cos(xy) \cdot (yx'+y'x)=1+y'$$.

Solve for \(\frac{dy}{dx}\)

Now, we rearrange the equation to solve for \(\frac{dy}{dx}\) (denoted as y'): $$yx' + yy'x = 1-y'$$. Transpose all terms with y' on one side, and the rest on the other side: $$yy'x + y' = 1 - yx'.$$ Factor y' from the left side to isolate it: $$y'(yx+1) = 1 - yx'.$$ Finally, divide by \((yx+1)\) to get the derivative \(\frac{dy}{dx}\): $$\frac{dy}{dx} = y' = \frac{1 - yx'}{yx+1}.$$

Key concepts

Chain Rule

In calculus, the chain rule is an important technique for differentiating compositions of functions. It is instrumental when dealing with implicit differentiation. The chain rule is used when a function is nested within another function, allowing us to differentiate the outer function first and then multiply it by the derivative of the inner function.

When applying the chain rule, it helps to remember:

• Differentiating an outer function, keeping the inner function intact.
• Multiplying the result by the derivative of the inner function.

In our exercise, while differentiating \( \sin(xy) \), \( xy \) is treated as a "nested" function, and the chain rule allows us to handle its differentiation efficiently. So, initially, you would differentiate \( \sin(u) = \cos(u) \) and then tackle the derivative of \( u = xy \) using additional rules for more complex functions.

Product Rule

The product rule is another essential tool in differentiating products of two functions. It's especially useful when dealing with expressions like \( xy \), where you have two variables multiplied together.

By the product rule, if you have two functions \( u(x) \) and \( v(x) \), their derivative is given by:

• \( (uv)' = u'v + uv' \)

In the expression \( xy \), \( x \) becomes one function and \( y \) becomes another. When applied, the product rule helps in breaking down the product into manageable parts. \( x \) remains constant, while differentiating \( y \) with respect to \( x \) calls in \( \frac{dy}{dx} \) - that's where the implicit differentiation shines.

Differentiation

Differentiation is the process in calculus by which we find the derivative of functions. Derivatives represent the rate of change of a function as its variable changes, and are key to understanding how things vary over time or space.

In mathematics, a derivative is a measure that shows how a function changes as its input changes. In simple terms, it tells us how fast or slow a function is moving at a particular point.
When applying differentiation in problems, it's important to understand:

• Basic rules, such as power and sum rules.
• Advanced techniques like the chain rule and product rule, for handling composite functions or products.

Our task uses implicit differentiation. This means differentiating an equation in terms of one variable while that variable is intertwined with another. Differentiation sets up the skeleton for calculus problem solving, assisting us in unpacking functions into their individual rates of change.

Calculus Problem Solving

Calculus problem solving involves breaking down complex mathematical tasks into manageable steps. With tools like the chain rule, product rule, and implicit differentiation, calculus becomes easier to tackle.

Effective problem solving in calculus involves:

• Understanding the problem thoroughly before finding solutions.
• Identifying which calculus rules or techniques apply to the situation at hand.
• Solving step-by-step while ensuring clear logical progression.

In our exercise, implicit differentiation allows us to unpack a trigonometric equation into a derivative. The process requires gentle attention to detail, recognizing each component's role, and carefully applying differentiation techniques step-by-step.
Constant practice and exposure to different types of calculus problems boost your problem-solving skills and confidence in handling complex equations.