Chapter 28: Problem 17
Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) by partial differentiation, when \(x \tan y=y \sin x\).
Short Answer
Expert verified
\( \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{y \cos x - \tan y}{x \sec^2 y - \sin x} \) is the derivative.
Step by step solution
01
Identify the function to differentiate
We are given the equation \( x \tan y = y \sin x \). Our task is to find the derivative \( \frac{\mathrm{d} y}{\mathrm{~d} x} \) using partial differentiation.
02
Differentiate both sides with respect to x
Take the derivative of both sides of the equation with respect to \( x \). On the left-hand side, using the product rule: \( \frac{\mathrm{d}}{\mathrm{d}x} (x \tan y) = x \frac{\mathrm{d}}{\mathrm{d}x} (\tan y) + \tan y \frac{\mathrm{d}}{\mathrm{d}x} (x) \). On the right-hand side, using the product rule: \( \frac{\mathrm{d}}{\mathrm{d}x} (y \sin x) = y \frac{\mathrm{d}}{\mathrm{d}x} (\sin x) + \sin x \frac{\mathrm{d}}{\mathrm{d}x} (y) \).
03
Apply the derivative formulas
For the left side: \( x \tan y \Rightarrow x \sec^2 y \frac{\mathrm{d}y}{\mathrm{d}x} + \tan y \). For the right side: \( y \sin x \Rightarrow y \cos x + \sin x \frac{\mathrm{d}y}{\mathrm{d}x} \).
04
Set the differentiated expressions equal to each other
After differentiating, set the expressions from both sides equal: \[ x \sec^2 y \frac{\mathrm{d}y}{\mathrm{d}x} + \tan y = y \cos x + \sin x \frac{\mathrm{d}y}{\mathrm{d}x} \].
05
Solve for \( \frac{\mathrm{d} y}{\mathrm{d} x} \)
Rearrange the equation from Step 4 to solve for \( \frac{\mathrm{d} y}{\mathrm{d} x} \). Move terms involving \( \frac{\mathrm{d} y}{\mathrm{d} x} \) to one side and factor them out:\[ x \sec^2 y \frac{\mathrm{d} y}{\mathrm{d} x} - \sin x \frac{\mathrm{d} y}{\mathrm{d} x} = y \cos x - \tan y \].Factor:\[ \frac{\mathrm{d} y}{\mathrm{d} x} (x \sec^2 y - \sin x) = y \cos x - \tan y \].Finally, solve for \( \frac{\mathrm{d} y}{\mathrm{d} x} \):\[ \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{y \cos x - \tan y}{x \sec^2 y - \sin x} \].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule
When differentiating a function, we often encounter products of two functions. In such cases, the **product rule** is a handy tool. It's a basic formula used in differentiation, which helps when a function is composed of two multiplied variables, each dependent on the differentiation variable.
Imagine you have a function given as the product of two functions, let's name them: \( u(x) \) and \( v(x) \). The product rule states that the derivative of their product with respect to \( x \) is:
Always remember:
Imagine you have a function given as the product of two functions, let's name them: \( u(x) \) and \( v(x) \). The product rule states that the derivative of their product with respect to \( x \) is:
- \( \frac{\mathrm{d}}{\mathrm{d}x} [u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)
Always remember:
- Keep one function constant while differentiating the other, then switch their roles.
- Add the results to get the total derivative of the product.
Trigonometric Functions
The world of calculus frequently involves **trigonometric functions**. They're foundational in many mathematical problems, including our current differentiation task. Familiarizing oneself with derivatives of basic trigonometric functions can make a significant difference in solving such problems smoothly.
Here's a quick review of the derivatives of basic trigonometric functions:
Practicing these derivatives and understanding how to apply them will simplify differentiation tasks with trigonometric functions involved.
Here's a quick review of the derivatives of basic trigonometric functions:
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( \cos x \) is \(-\sin x \).
- The derivative of \( \tan x \) is \( \sec^2 x \).
Practicing these derivatives and understanding how to apply them will simplify differentiation tasks with trigonometric functions involved.
Chain Rule
In differentiation, the **chain rule** is extremely useful for solving functions nested within other functions. It's especially pivotal when dealing with composite functions where one function sits "inside" another.
For example, you have \( \tan y \) in the given problem, where \( y \) itself is a function of \( x \). This is where the chain rule comes into play. If \( y = f(x) \), then when finding the derivative of \( \tan y \) with respect to \( x \), you chain the derivative of the outside function (\( \tan y \)) with the inside function's derivative (\( y \)).
The chain rule formula is:
For instance, the derivative of \( \tan y \) with respect to \( x \) is \( \sec^2 y \cdot \frac{\mathrm{d}y}{\mathrm{d}x} \). You first differentiate \( \tan y \) as if \( y \) were a simple variable, and then multiply by the derivative of \( y \) with respect to \( x \).
Mastering the chain rule will make working with partially differentiated equations much easier and more intuitive.
For example, you have \( \tan y \) in the given problem, where \( y \) itself is a function of \( x \). This is where the chain rule comes into play. If \( y = f(x) \), then when finding the derivative of \( \tan y \) with respect to \( x \), you chain the derivative of the outside function (\( \tan y \)) with the inside function's derivative (\( y \)).
The chain rule formula is:
- \( \frac{\mathrm{d}}{\mathrm{d}x} [f(g(x))] = f'(g(x)) \cdot g'(x) \)
For instance, the derivative of \( \tan y \) with respect to \( x \) is \( \sec^2 y \cdot \frac{\mathrm{d}y}{\mathrm{d}x} \). You first differentiate \( \tan y \) as if \( y \) were a simple variable, and then multiply by the derivative of \( y \) with respect to \( x \).
Mastering the chain rule will make working with partially differentiated equations much easier and more intuitive.