Problem 10
Question
(x) \(d\left(\tan ^{-1} \frac{y}{x}\right)=\frac{x d y-y d x}{x^{2}+y^{2}}\)
Step-by-Step Solution
Verified Answer
Differentiating, we confirm: \(d(\tan^{-1} (\frac{y}{x})) = \frac{x \, dy - y \, dx}{x^2 + y^2}\).
1Step 1: Understand the Problem Statement
The problem asks us to differentiate the inverse tangent function, \(\tan^{-1} \left(\frac{y}{x}\right)\), and show that its differential can be expressed as \(\frac{x \, dy - y \, dx}{x^2 + y^2}\).
2Step 2: Apply the Chain Rule
The function \(z = \tan^{-1} \left(\frac{y}{x}\right)\) can be seen as a composite function: \(u = \frac{y}{x} \) and \(z = \tan^{-1}(u)\). Using the chain rule: \(d z = \frac{d}{du}(\tan^{-1}(u)) \cdot du\).
3Step 3: Differentiate \(\tan^{-1}(u)\)
The derivative of \(\tan^{-1}(u)\) with respect to \(u\) is \( \frac{1}{1+u^2}\). So, \(d z = \frac{1}{1+u^2} \cdot du\).
4Step 4: Differentiate \(u = \frac{y}{x}\) Using the Quotient Rule
The derivative \(du\) is \(\frac{x \, dy - y \, dx}{x^2}\) using the quotient rule \(\frac{d}{dx}\left(\frac{y}{x}\right) = \frac{x \, d(y/x) - y \, d(1)}{x^2}\).
5Step 5: Substitute \(du\) and Simplify
Substitute \(du\) into the expression for \(dz\):\[{dz = \frac{1}{1+\left(\frac{y}{x}\right)^2} * \frac{x \, dy - y \, dx}{x^2}} = \frac{x^2}{x^2+y^2} \cdot \frac{x \, dy - y \, dx}{x^2} = \frac{x \, dy - y \, dx}{x^2 + y^2}.\]
Key Concepts
Chain RuleQuotient RuleInverse Trigonometric Functions
Chain Rule
When dealing with composite functions in calculus, the Chain Rule becomes an indispensable tool. It helps us differentiate functions that are "inside" other functions. Think of it as peeling an onion, addressing each layer one by one.
A composite function may look like this: if we have a function, say \(z = f(g(x))\), the Chain Rule tells us that the derivative \(dz\) is the product of the derivative of the outer function evaluated at the inner function, and the derivative of the inner function itself. Therefore, \(dz = f'(g(x)) \cdot g'(x)\).
In our exercise, we see \(z = \tan^{-1}(u)\) with \(u = \frac{y}{x}\). We've separated the stages into two derivatives: first applying the rule to the outside function, \(\tan^{-1}\), and then to the inner function, \(\frac{y}{x}\).
A composite function may look like this: if we have a function, say \(z = f(g(x))\), the Chain Rule tells us that the derivative \(dz\) is the product of the derivative of the outer function evaluated at the inner function, and the derivative of the inner function itself. Therefore, \(dz = f'(g(x)) \cdot g'(x)\).
In our exercise, we see \(z = \tan^{-1}(u)\) with \(u = \frac{y}{x}\). We've separated the stages into two derivatives: first applying the rule to the outside function, \(\tan^{-1}\), and then to the inner function, \(\frac{y}{x}\).
- The derivative of \(\tan^{-1}(u)\) with respect to \(u\) is \(\frac{1}{1+u^2}\).
- Overall, \(dz\) is formed by multiplying this answer by the derivative of \(u\), which we need to find separately.
Quotient Rule
When finding the derivative of a quotient of two functions, the Quotient Rule comes into play. This rule is especially useful when our function takes the form \(\frac{f(x)}{g(x)}\). The formula for this rule is slightly more intricate than the product rule, ensuring we correctly consider each part of the fraction.
The Quotient Rule states that the derivative \(\frac{d}{dx}\left(\frac{f}{g}\right)\) is given by \(\frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}\). It’s a way to tackle fractions while maintaining precision.
In our situation, when differentiating \(u = \frac{y}{x}\), we apply this rule directly:
The Quotient Rule states that the derivative \(\frac{d}{dx}\left(\frac{f}{g}\right)\) is given by \(\frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}\). It’s a way to tackle fractions while maintaining precision.
In our situation, when differentiating \(u = \frac{y}{x}\), we apply this rule directly:
- The result is \(\frac{x \cdot dy - y \cdot dx}{x^2}\).
Inverse Trigonometric Functions
Inverse trigonometric functions, such as \(\tan^{-1}(x)\), also called arctangent, offer a way to navigate relationships in right-angled triangles that go beyond direct angles. These functions reverse the trigonometric functions, hence the term "inverse."
With \(\tan^{-1}(x)\), it maps a ratio back to an angle, converting a tangent value into the corresponding angle measure. This type of function is crucial when solving for angles in applied math and physics.
Unlike straightforward sin, cos, or tan functions, inverse functions have special derivatives that require careful application when differentiating. The derivative of \(\tan^{-1}(u)\) specifically is \(\frac{1}{1 + u^2}\) — crucial knowledge for tackling problems like our exercise, where this derivative forms the backbone of the Chain Rule application.
With \(\tan^{-1}(x)\), it maps a ratio back to an angle, converting a tangent value into the corresponding angle measure. This type of function is crucial when solving for angles in applied math and physics.
Unlike straightforward sin, cos, or tan functions, inverse functions have special derivatives that require careful application when differentiating. The derivative of \(\tan^{-1}(u)\) specifically is \(\frac{1}{1 + u^2}\) — crucial knowledge for tackling problems like our exercise, where this derivative forms the backbone of the Chain Rule application.
- These derivatives can often look unusual or unintuitive, but they are incredibly useful for expressing the rate of change of these inverse relationships.
Other exercises in this chapter
Problem 8
(viii) \(d\left(\frac{1}{2} \log \left(x^{2}+y^{2}\right)\right)=\frac{x d x-y d y}{x^{2}+y^{2}}\)
View solution Problem 9
(ix) \(d\left(\frac{1}{2} \log \frac{x+y}{x-y}\right)=\frac{x d y-y d x}{x^{2}-y^{2}}\)
View solution Problem 13
(xiii) \(d\left(\frac{x^{2}}{y}\right)=\frac{2 y x d x-x^{2} d y}{y^{2}}\)
View solution Problem 14
(xiv) \(d\left(\frac{y^{2}}{x}\right)=\frac{2 x y d y-y^{2} d x}{x^{2}}\)
View solution