91Ó°ÊÓ

When a function is composed of two or more functions multiplied together, like the term \(2xy\) in our equation, we need to use the product rule to differentiate it. The product rule is a fundamental tool in calculus, especially for implicit differentiation. It states that the derivative of a product of two functions \(u\) and \(v\) is given by:

• \((u'v) + (uv')\)

Here, \(u\) could represent the function \(2x\) and \(v\) could be \(y\). This means we differentiate \(2x\) and multiply it by \(y\), and then differentiate \(y\) and multiply it by \(2x\). During implicit differentiation, the derivative of \(y\) with respect to \(x\) is \(dy/dx\). Adding the derivatives together, we apply the rule as follows:

• Differentiate \(2x\) to get 2.
• Multiply by \(y\).
• Then differentiate \(y\) to get \(dy/dx\).
• And multiply it by \(2x\).

Putting this together following the product rule gives us \(2(y + x \cdot \frac{dy}{dx})\), which is essential to correctly differentiate terms involving both \(x\) and \(y\).
Remember that the product rule helps keep track of both parts of the product, especially vital when dealing with implicit equations.

While differentiating constant terms, remember that the derivative of any constant is zero. Constants are values that do not change, so their slope along any curve is always flat. In our example, the \(+3\) in the equation \(2xy + 3 = 0\) is such a constant.Differentiating \(+3\) leads to zero because it does not vary with \(x\) or \(y\). Thus, it vanishes during differentiation. You simply write

• \(\frac{d}{dx}(3) = 0\)

Similarly, anytime you encounter a constant in differentiation tasks, you can effortlessly set its derivative to zero and move forward with the remaining parts of the expression.Handling constants correctly is essential in simplifying equations after differentiation. It ensures that no extra terms are carried around unnecessarily in the process of solving for \(dy/dx\).

After successfully differentiating the equation and applying rules like the product rule and constants differentiation, you'll notice terms involving \(dy/dx\) need to be isolated to solve for it.From the differentiated equation \(2(y + x \cdot \frac{dy}{dx}) = 0\), the goal is to develop an expression solely in terms of \(dy/dx\). Hence, you rewrite it as \(2x \cdot \frac{dy}{dx} = -2y\) by shifting all terms without \(dy/dx\) to the opposite side.

• This step requires moving **usual** rearranging techniques, just like solving algebraic equations.
• Always aim to bring like terms on one side to simplify computations.

In this exercise, dividing through by \(2x\) yields \(\frac{dy}{dx} = -\frac{y}{x}\). The simple algebraic manipulation of swapping terms while keeping track of signs is vital when isolating \(dy/dx\).Effectively isolating \(dy/dx\) allows you to express the derivative cleanly and clearly, showcasing the relationship between the derivatives and variables involved.