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Find \(d y / d x\) in terms of \(x\) and \(y\). $$ \sin (x y)+y=y \cos x $$

Short Answer

Expert verified
After simplifying, \(d y / d x\) in terms of \(x\) and \(y\) is \((y \sin x - x y \cos(xy) ) / ( \cos x - x \cos(xy) -1)\)

Step by step solution

01

Differentiate Both Sides

Differentiate both sides of the equation with respect to \(x\). Remember, when performing differentiation involving products like \(x\) and \(y\), use the product rule. When differentiating \(y\) with respect to \(x\), use dy/dx (the derivative of \(y\) with respect to \(x\) must be provided in the final result).
02

Apply the Product Rule

Apply the product rule which is \(uv' + vu'\) where u and v are functions of \(x\). So differentiating \(\sin(x y)\) gives us \(\cos(xy) * (y + xd y / dx)\). Differentiating \(y\) on the left side gives \(d y / dx\).
03

Differentiate Y Cos X

Differentiating \(y \cos(x)\) on the right side involves using the product rule again. The derivative of this part is \((d y / dx) \cos(x) - y \sin(x)\).
04

Combine All Parts

Combine all of these differentiated parts to form the equation: \(\cos(xy) * (y + xd y / dx) + d y / dx = (d y / dx) \cos(x) - y \sin(x)\).
05

Isolate Dy/Dx

Rearrange the equation to isolate dy/dx on one side. We obtain dy/dx on the left side of the equation, and the rest of the terms on the right side.
06

Simplify the Result

Simplify the rest of the terms on the right side to obtain the final expression for dy/dx.

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Key Concepts

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

Derivative Rules
When working with equations involving functions and derivatives, understanding derivative rules is crucial. Derivatives capture the rate of change of a function in relation to one variable, typically time or space.
In calculus, these rules help us determine how functions behave by simplifying the differentiation process. The most common rule is the power rule, which states that the derivative of a power function is reduced by one order. There is also the chain rule, which is used for composite functions.
However, in this instance, we are focusing on implicit differentiation, where the derivative rules also include treating respect to one variable while considering the interdependence of another variable, such as the function of both x and y. Here, d/dx is used while differentiating y in relation to x, allowing us to find dy/dx when y is defined implicitly in terms of x.
Product Rule
The product rule is a fundamental differentiation technique necessary when dealing with products of functions. It allows us to differentiate a function that is the product of two or more functions.
For two functions u and v, the product rule can be written as:\[ (uv)' = u'v + uv' \] Applying this involves taking the derivative of the first function and multiplying it by the second, then adding the product of the first function and the derivative of the second function.
In our problem, we see this applied in differentiating \( \sin(xy) \) and \( y \cos(x) \), requiring us to meticulously apply the product rule to find each respective derivative. This ensures that all terms emerge properly when solving for the derivative with respect to x.
Trigonometric Functions
Trigonometric functions often appear in calculus problems and add an additional layer of complexity to differentiation. These functions include sine, cosine, and tangent, among others.
Each trigonometric function has a specific derivative: for instance, \( \frac{d}{dx}[\sin(x)] = \cos(x) \) and \( \frac{d}{dx}[\cos(x)] = -\sin(x) \). Knowing these derivatives allows us to solve problems involving trigonometric expressions efficiently.
In the given problem, we apply these derivatives while being mindful of the product rule and chain rule, especially within compound expressions involving terms like \( \sin(xy) \) and \( y\cos(x) \). Differentiating trigonometric expressions requires a careful blend of these fundamental functions with respect to their interdependent variables.
Differentiation Techniques
Differentiation techniques encompass various methods to find the derivative of a function, crucial for revealing hidden behaviors in complex expressions.
In this exercise, implicit differentiation is the technique used. Unlike standard differentiation, implicit differentiation allows differentiation when one variable cannot be explicitly separated from another. Each term in the expression is differentiated in terms of x, treating y as an implicit function of x.
We also encounter different rules, such as the chain rule when differentiating composite functions, and the product rule when we deal with products of functions involving x and y. Comprehensive understanding and careful application of these methods allow us to arrive at the derivative \( \frac{dy}{dx} \) in terms of x and y, leading to a solution for complex equations involving several variables.

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Most popular questions from this chapter

A wheel of radius 5 meters is oriented vertically and spinning counterclockwise at a rate of 7 revolutions per minute. If the origin is placed at the center of the wheel a point on the rim has a horizontal position of \((7,0)\) at time \(t=0\). What is the horizontal component of the point's velocity at \(t=2 ?\)

Approximating the function \(f(x)=\sin x\) near \(x=0\) by using polynomials. The point of this problem is to show you how the values of \(\sin x\) can be approximated numerically with a very high degree of accuracy. It is an introduction to Taylor polynomials. (a) Find the equation of the line tangent to \(f(x)=\sin x\) at \(x=0\). (b) Find the equation of a quadratic \(Q(x)=a+b x+c x^{2}\) such that the function \(Q(x)\) and its nonzero derivatives match those of \(\sin x\) at \(x=0\). In other words, \(Q(0)=f(0), Q^{\prime}(0)=f^{\prime}(0)\), and \(Q^{\prime \prime}(0)=f^{\prime \prime}(0)\). The quadratic that you found is the quadratic that best "fits" the sine curve near \(x=0\). In fact, the "quadratic" turns out not to really be a quadratic at all. Sine is an odd function, so there is no parabola that "fits" the sine curve well at \(x=0\). (c) Find the equation of a cubic \(C(x)=a+b x+c x^{2}+d x^{3}\) such that the function \(C(x)\) and its nonzero derivatives match those of \(\sin x\) at \(x=0 .\) In other words, \(C(0)=f(0), C^{\prime}(0)=f^{\prime}(0), C^{\prime \prime}(0)=f^{\prime \prime}(0)\), and \(C^{\prime \prime \prime}(0)=f^{\prime \prime \prime}(0) .\) The cubic that you found is the cubic that best "fits" the sine curve near \(x=0\). Using a calculator, on the same set of axes graph \(\sin x\), the tangent line to \(\sin x\) at \(x=0\), and \(C(x)\), the cubic you found. Now "zoom in" around \(x=0 .\) Can you see that near \(x=0\) the line is a good fit to the sine curve but the cubic is an even better fit for small \(x\) and the cubic hugs the sine curve for longer? The next set of questions asks you to investigate how good the fit is. (d) Use \(C(x)\) from part (c) to estimate the following, and then compare with the actual value using a calculator. \(\begin{array}{lllll}\sin (0.01) & \sin (0.1) & \sin (0.5) & \sin (1) & \sin (3)\end{array}\) (e) (Challenge) Find the "best" fifth degree polynomial approximation of \(\sin (x)\) for \(x\) near 0 by making sure that the first five derivatives of the polynomial match those of \(\sin (x)\) when evaluated at \(x=0 .\) Graph \(\sin x\) along with its first, third, and fifth degree polynomial approximations on your graphing calculator. The higher the degree of the polynomial, the better the fit to \(\sin x\) near \(x=0\), right? (f) Using a calculator, on the same set of axes graph \(\sin x\) and the polynomial given below. $$x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}+\frac{x^{9}}{362880}$$ This polynomial is an even better fit than the last one, right? Now graph the difference between \(\sin x\) and this polynomial; in other words, graph $$y=\sin x-\left[x-\frac{x^{3}}{6}+\frac{x^{5}}{120}-\frac{x^{7}}{5040}+\frac{x^{9}}{362880}\right]$$ On approximately what interval is the difference between \(\sin x\) and this polynomial less than \(0.005\) ?

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