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Differentiate implicitly to find \(d y / d x\). $$ \frac{x}{x^{2}+y^{2}}-y^{2}=6 $$

Short Answer

Expert verified
The derivative of the given equation with respect to x, \( dy / dx\), is \((x^{2} / y - (x^{2} + y^{2})) / (2x - 2)\)

Step by step solution

01

Differentiate both sides of the equation with respect to \(x\)

Differentiating \(x/(x^{2}+y^{2})\) using the quotient rule, gives \((x^{2}+y^{2} - 2x^2 - 2xy * dy/dx) / (x^{2}+y^{2})^2\). Differentiating \(-y^{2}\) gives \(-2y * dy/dx\). Differentiating 6 on the right side gives 0 as the derivative of a constant is zero.
02

Simplify and bring terms involving \(dy/dx\) on one side

Bringing all terms with dy/dx to the left gives \((2x^2 + 2xy * dy/dx - 2y * dy/dx) / (x^{2}+y^{2})^2 = x /(x^{2}+y^{2}) - 6\). Thus, \( dy/dx(2xy - 2y) = x^{2} - (x^{2} + y^{2})^2 \).
03

Factor out and simplify

Factor out dy/dx to get \( dy/dx = (x^{2} - (x^{2} + y^{2})^2) / (2xy - 2y)\). Simplifying further gives \( dy/dx = (x^{2} / y - (x^{2} + y^{2})) / (2x - 2)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quotient Rule
In calculus, the quotient rule is a technique for finding the derivative of a function that is the ratio of two differentiable functions. It states that if you have a function \( u(x) / v(x) \), where both \( u \) and \( v \) are differentiable, the derivative \( u'/v' \) is given by:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]
Applying this rule, as seen in the given exercise, allows you to differentiate complex fractions systematically. In our example, we differentiate the numerator and the denominator separately, then use the formula above to combine these derivatives, keeping track of the minus sign that appears when applying the rule. It's important to keep your work organized to avoid mistakes with signs or the sequence of the terms.
Differentiating Constants
When it comes to taking derivatives, constants present a unique but straightforward case. In differentiation, any constant term \( c \) has a derivative of zero. This is because a constant does not change as the variable changes—it has no rate of change, which is essentially what a derivative measures.
In the exercise provided, the term 6 on the right side of the equation is a constant. So when we differentiate this side of the equation with respect to \( x \), it becomes 0. Understanding this principle helps in simplifying equations when taking derivatives, as constant terms vanish upon differentiation. Remembering this rule can save time and reduce complexity when solving calculus problems.
Simplifying Derivatives
Once you've applied the rules of differentiation, it's often necessary to simplify the derivative expression to make it more manageable and to reveal its characteristics more clearly. Simplifying may include combining like terms, factoring, and canceling common factors.
In our example, after applying the quotient rule and differentiating all terms, we collect like terms and simplify the resulting expression. Simplifying expressions often makes it easier to see the relationship between variables and to continue with further manipulations, such as solving for \( dy/dx \). Remember, the goal of simplification is to make your derivative as clear and concise as possible without losing any essential information.
Factor Out Derivatives
Factoring out involves grouping together terms that share a common factor to simplify an expression or equation. In the context of differentiation, after taking the derivative, if there are multiple terms that contain the derivative of the function (e.g., \( dy/dx \) in implicit differentiation), you can factor out \( dy/dx \) to isolate it and solve for the derivative explicitly.
This is particularly useful in implicit differentiation, as seen in the exercise. We collect all terms with \( dy/dx \) on one side and factor it out. This leaves us with an equation where \( dy/dx \) is by itself on one side, making it straightforward to solve for \( dy/dx \). Factoring is a powerful tool in simplifying complex expressions and is essential for finding the derivative of a function with respect to one of its variables.

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Most popular questions from this chapter

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