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Magazine Chapter 2: Problem 16
Find \(d y / d x\) by implicit differentiation. $$\tan (x-y)=\frac{y}{1+x^{2}}$$
Short Answer
Expert verified
\(\frac{dy}{dx} = \frac{\sec^2(x-y)\cdot(1+x^2)^2 + 2xy}{(1+x^2)\cdot\sec^2(x-y) + 1+x^2}\)
Step by step solution
01
Differentiate Both Sides
To differentiate both sides with respect to \(x\), apply implicit differentiation. On the left, differentiate \(\tan(x-y)\) using the chain rule which results in \(\sec^2(x-y)\cdot(1 - \frac{dy}{dx})\). On the right, differentiate \(\frac{y}{1+x^2}\) using the quotient rule resulting in \(\frac{(1+x^2)\cdot \frac{dy}{dx} - y\cdot 2x}{(1+x^2)^2}\).
02
Set Up the Equation
Equate the derivatives from both sides: \[\sec^2(x-y) \cdot (1 - \frac{dy}{dx}) = \frac{(1+x^2)\cdot \frac{dy}{dx} - y\cdot 2x}{(1+x^2)^2}.\] This represents the differentiated equation.
03
Separate \(\frac{dy}{dx}\) Terms
Rearrange the equation to gather all terms involving \(\frac{dy}{dx}\) to one side of the equation: \[ \sec^2(x-y) - \sec^2(x-y)\cdot\frac{dy}{dx} = \frac{dy}{dx}\cdot\frac{1+x^2}{(1+x^2)^2} - \frac{2xy}{(1+x^2)^2}. \]
04
Isolate \(\frac{dy}{dx}\)
Factor out \(\frac{dy}{dx}\) from the terms on the right-hand side and solve for it: \[\frac{dy}{dx}\left(\frac{1+x^2}{(1+x^2)^2} + \sec^2(x-y)\right) = \sec^2(x-y) + \frac{2xy}{(1+x^2)^2}.\] Then, \(\frac{dy}{dx}\) is isolated: \[\frac{dy}{dx} = \frac{\sec^2(x-y) + \frac{2xy}{(1+x^2)^2}}{\frac{1+x^2}{(1+x^2)^2} + \sec^2(x-y)}.\]
05
Simplify the Expression
Combine terms and simplify the expression for \(\frac{dy}{dx}\) as needed: \[\frac{dy}{dx} = \frac{\sec^2(x-y)\cdot(1+x^2)^2 + 2xy}{(1+x^2)\cdot\sec^2(x-y) + 1+x^2}.\] This is your final solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
The Chain Rule
The chain rule is a fundamental tool in calculus used when differentiating composite functions. It simplifies the process of finding the derivative of functions within other functions. In essence, the chain rule helps you breakdown complex derivatives into manageable parts by applying differentiation to both the outer and inner functions.
Consider a function of the form \(f(g(x))\). To differentiate this, you would take the derivative of \(f\) with respect to \(g(x)\), written as \(f'(g(x))\), and then multiply it by the derivative of \(g(x)\) with respect to \(x\), noted as \(g'(x)\). This gives us \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).
Consider a function of the form \(f(g(x))\). To differentiate this, you would take the derivative of \(f\) with respect to \(g(x)\), written as \(f'(g(x))\), and then multiply it by the derivative of \(g(x)\) with respect to \(x\), noted as \(g'(x)\). This gives us \((f(g(x)))' = f'(g(x)) \cdot g'(x)\).
- For example, differentiating \(\tan(x-y)\) involves recognizing \(x-y\) as the inner function and \(\tan\) as the outer function.
- Using the chain rule here gives you \(\sec^2(x-y)\cdot(1 - \frac{dy}{dx})\), as you differentiate \(x-y\) as part of the chain process.
The Quotient Rule
The quotient rule comes into play when you need to differentiate a function formed by the division of two different functions. Understanding and applying this rule correctly is essential for working with rational functions and their derivatives.
When you have a function \(\frac{g(x)}{h(x)}\), the quotient rule states that the derivative of the function is given by:
When you have a function \(\frac{g(x)}{h(x)}\), the quotient rule states that the derivative of the function is given by:
- \((\frac{g}{h})' = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}\)
- Numerator: \((1+x^2) \cdot \frac{dy}{dx} - y \cdot 2x\)
- Denominator: \((1+x^2)^2\)
Trigonometric Differentiation
Trigonometric differentiation deals with finding the derivatives of trigonometric functions. It's a critical skill, particularly in calculus problems like the given exercise, where trigonometric functions are present on one side of the equation.
For the function \(\tan(x-y)\), its derivative can be found using a trigonometric identity where the derivative of \(\tan(u)\) is \(\sec^2(u)\). Combined with the chain rule, it becomes:
For the function \(\tan(x-y)\), its derivative can be found using a trigonometric identity where the derivative of \(\tan(u)\) is \(\sec^2(u)\). Combined with the chain rule, it becomes:
- \(\sec^2(x-y)\cdot(1 - \frac{dy}{dx})\)
Calculus Problem Solving
Solving calculus problems often involves a systematic approach. Key skills include applying rules like the chain and quotient rules, understanding trigonometric derivatives, and algebraic manipulation for simplification.
For the exercise provided, implicit differentiation is essential due to the presence of both \(x\) and \(y\) in the function \(\tan(x-y) = \frac{y}{1+x^2}\). Implicit differentiation allows you to find \(\frac{dy}{dx}\) by treating \(y\) as a function of \(x\) even though \(y\) isn't isolated from \(x\).
For the exercise provided, implicit differentiation is essential due to the presence of both \(x\) and \(y\) in the function \(\tan(x-y) = \frac{y}{1+x^2}\). Implicit differentiation allows you to find \(\frac{dy}{dx}\) by treating \(y\) as a function of \(x\) even though \(y\) isn't isolated from \(x\).
- Start by differentiating each side of the equation with respect to \(x\).
- Collect terms involving \(\frac{dy}{dx}\) to one side to isolate the derivative.
- Factor out \(\frac{dy}{dx}\), and then finally solve for it.
