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Implicit differentiation Use implicit differentiation to find \(\frac{d y}{d x}\). $$\sin x y=x+y$$

Short Answer

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Question: Find the derivative of \(y\) with respect to \(x\) for the given equation: \(\sin xy = x + y\). Answer: \(\frac{dy}{dx} = \frac{1 - y \cos(xy)}{x \cos(xy) - 1}\)

Step by step solution

01

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

Differentiate \(\sin xy = x + y\) with respect to \(x\): $$\frac{d}{dx}(\sin xy) = \frac{d}{dx}(x + y)$$
02

Apply the chain rule to expressions involving \(y\)

In the left side of the equation, we need to apply the chain rule since there is "x y" inside sin function. For expressing it correctly we will use d as the differential. Next, differentiate \(x+y\): $$\cos(xy) (\frac{d}{dx}(x y)) = 1 + \frac{dy}{dx}$$
03

Apply the product rule to the expression inside sin

Apply the product rule to the differentiation of \(x y\): $$\cos(xy) (\frac{d}{dx}(x) y + x \frac{d}{dx}(y)) = 1 + \frac{dy}{dx}$$
04

Substitute the differentiation results

Now we differentiate x and y: $$\cos(xy) (y + x \frac{dy}{dx}) = 1 + \frac{dy}{dx}$$
05

Solve the equation for \(\frac{dy}{dx}\)

Firstly, organize the equation to isolate dy/dx: $$x \cos(xy) \frac{dy}{dx} - \frac{dy}{dx} = 1 - y \cos(xy)$$ Then, factor \(\frac{dy}{dx}\): $$\frac{dy}{dx}(x \cos(xy) - 1) = 1 - y \cos(xy)$$ Now, divide both sides of the equation by \((x \cos(xy) - 1)\): $$\frac{dy}{dx} = \frac{1 - y \cos(xy)}{x \cos(xy) - 1}$$ Therefore, the derivative of \(y\) with respect to \(x\) is: $$\frac{dy}{dx} = \frac{1 - y \cos(xy)}{x \cos(xy) - 1}$$

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Key Concepts

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

Chain Rule
To understand implicit differentiation, we must grasp the concept of the Chain Rule. The Chain Rule is essential whenever you're differentiating compositions of functions. In simple terms, it helps you differentiate a function that is nested within another function.
When we have an equation like \( \sin xy = x + y \), we notice "\( xy \)" is inside the sine function, which means we need to apply the Chain Rule.
Here's a quick breakdown: In our given equation, the outer function is \( \sin(u) \) and the inner function \( u = xy \). When differentiating \( \sin u \) with respect to \( x \), the chain rule requires us to multiply the derivative of the outer function \( \cos(u) \) by the derivative of the inner function \( \frac{du}{dx} \).
Thus, we get \[ \cos(xy) \cdot \frac{d(xy)}{dx} \]. It’s a way to 'chain' different parts of a composite function together for derivation purposes.
This technique is powerful when working with implicit differentiation where functions of \( y \) are involved.
Product Rule
The Product Rule is another key tool in differentiation, particularly useful when dealing with products of functions. In the context of our exercise, we find \( xy \) within our equation \( \sin xy = x + y \). Here, \( xy \) is a product of \( x \) and \( y \), which requires us to use the Product Rule.
The Product Rule states that when differentiating a product of two functions, say \( u(x) \cdot v(x) \), the formula is \( u'v + uv' \).
Applying this to \( x \cdot y \), gives \( \frac{d}{dx}(xy) = \frac{d}{dx}(x)y + x\frac{d}{dx}(y) = y + x\frac{dy}{dx} \).
This derivative then feeds into the chain rule from the previous section, allowing us to further differentiate more complex parts of the equation.
The Product Rule makes finding derivatives of multiplicative expressions straightforward and systematic, especially when dealing with terms that involve products of variables.
Differentiation Techniques
Differentiation is a staple method in calculus used to find the rate at which one quantity changes with respect to another. There are numerous techniques for differentiation, especially useful for finding derivatives in implicit and explicit forms.
Some key techniques used here include the Chain Rule and Product Rule. However, implicit differentiation particularly shines when dealing with functions not easily solved for one variable. When \( y \) is not isolated, like in our example \( \sin xy = x + y \), implicit differentiation allows us to differentiate both sides with respect to \( x \) without explicitly solving for \( y \).
Start by differentiating every term with respect to \( x \). Then, whenever \( y \) appears, multiply its derivative by \( \frac{dy}{dx} \) since \( y \) is considered a function of \( x \).
Finally, after using these techniques, we often end up with a mix of \( y \) and \( \frac{dy}{dx} \). The next step is to rearrange to solve for \( \frac{dy}{dx} \), giving us the needed derivative even when \( y \) is not separately defined.
Implicit differentiation and its associated techniques make handling complex equations much more manageable.

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Most popular questions from this chapter

A particle travels clockwise on a circular path of diameter \(R,\) monitored by a sensor on the circle at point \(P ;\) the other endpoint of the diameter on which the sensor lies is \(Q\) (see figure). Let \(\theta\) be the angle between the diameter \(P Q\) and the line from the sensor to the particle. Let \(c\) be the length of the chord from the particle's position to \(Q\) a. Calculate \(d \theta / d c\) b. Evaluate \(\left.\frac{d \theta}{d c}\right|_{c=0}\)

Suppose \(y=L(x)=a x+b\) (with \(a \neq 0\) ) is the equation of the line tangent to the graph of a one-to-one function \(f\) at \(\left(x_{0}, y_{0}\right) .\) Also, suppose that \(y=M(x)=c x+d\) is the equation of the line tangent to the graph of \(f^{-1}\) at \(\left(y_{0}, x_{0}\right)\) a. Express \(a\) and \(b\) in terms of \(x_{0}\) and \(y_{0}\) b. Express \(c\) in terms of \(a,\) and \(d\) in terms of \(a, x_{0},\) and \(y_{0}\) c. Prove that \(L^{-1}(x)=M(x)\)

Find the derivative of the inverse of the following functions at the specified point on the graph of the inverse function. You do not need to find \(f^{-1}\) $$f(x)=x^{2}+1, \text { for } x \geq 0 ;(5,2)$$

Derivatives and inverse functions Suppose the slope of the curve \(y=f^{-1}(x)\) at (4,7) is \(\frac{4}{5}\) Find \(f^{\prime}(7)\)

Gravitational force The magnitude of the gravitational force between two objects of mass \(M\) and \(m\) is given by \(F(x)=-\frac{G M m}{x^{2}},\) where \(x\) is the distance between the centers of mass of the objects and \(G=6.7 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}\) is the gravitational constant (N stands for newton, the unit of force; the negative sign indicates an attractive force). a. Find the instantaneous rate of change of the force with respect to the distance between the objects. b. For two identical objects of mass \(M=m=0.1 \mathrm{kg},\) what is the instantaneous rate of change of the force at a separation of \(x=0.01 \mathrm{m} ?\) c. Does the instantaneous rate of change of the force increase or decrease with the separation? Explain.

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