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Chapter 3: Problem 36

Use implicit differentiation to find\(\frac{d y}{d x}.\) $$(x y+1)^{3}=x-y^{2}+8$$

Short Answer

Expert verified
Question: Find the derivative of the function with respect to x, given the equation (xy + 1)^3 = x - y^2 + 8. Answer: The derivative of the function with respect to x is given by: $$\frac{d(y)}{d x} = \frac{1 - 3(x y + 1)^2 y}{3(x y + 1)^2 x + 2y}$$

Step by step solution

01

Differentiate both sides of the equation with respect to x

First, we will take the derivative with respect to \(x\) for both sides of the equation: $$(x y+1)^{3}=x-y^{2}+8$$ Differentiating the left side, we use the chain rule and get: $$(3(x y + 1)^2)(\frac{d(xy+1)}{dx}) = \frac{d(x)}{dx} - \frac{d(y^{2})}{dx} + \frac{d(8)}{dx}$$
02

Differentiate the inner terms

Next, we apply the product rule in the left side of the equation (for the term \(xy+1\)). We also differentiate the terms on the right side: $$3(x y + 1)^2 (\frac{d(x y)}{d x} + \frac{d(1)}{d x}) = 1 - 2y \frac{d(y)}{d x} + 0$$
03

Continue applying the product rule

Now, to differentiate \(d(xy)\), we will use the product rule: $$\frac{d(x y)}{d x} = y \frac{d(x)}{d x} + x \frac{d(y)}{d x}$$ We will substitute this back into our previous expression: $$3(x y + 1)^2 (y + x \frac{d(y)}{d x}) = 1 - 2y \frac{d(y)}{d x}$$
04

Distribute the terms

Now, distribute the terms inside the parentheses and combine the terms with \(\frac{d(y)}{d x}\): $$3(x y + 1)^2 y + 3(x y + 1)^2 x \frac{d(y)}{d x} = 1 - 2y \frac{d(y)}{d x}$$
05

Isolate the term with \(\frac{dy}{dx}\)

Move all the terms containing \(\frac{d(y)}{d x}\) to one side of the equation and isolate it: $$3(x y + 1)^2 x \frac{d(y)}{d x} + 2y\frac{d(y)}{d x} = 1 - 3(x y + 1)^2 y$$ Now, we factor out \(\frac{d(y)}{d x}\) from the left side and isolate it: $$\frac{d(y)}{d x}(3(x y + 1)^2 x + 2y) = 1 - 3(x y + 1)^2 y$$ Finally, we solve for \(\frac{d(y)}{d x}\): $$\frac{d(y)}{d x} = \frac{1 - 3(x y + 1)^2 y}{3(x y + 1)^2 x + 2y}$$

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Key Concepts

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

Chain Rule
When dealing with functions embedded inside other functions, the chain rule becomes essential. It's like peeling an onion, where you find each layer influencing the next. The chain rule states that the derivative of a composite function can be found by taking the derivative of the outer function and then multiplying it by the derivative of the inner function.
For example, with \((x y + 1)^3\), the outer function is the cube function, and the inner function is \(x y + 1\).
  • Differentiating the outer layer: \(3(x y + 1)^2\)
  • Then, multiply by the derivative of the inner layer: \(\frac{d(x y + 1)}{dx}\)
This sequence of steps ensures all embedded changes are accounted for systematically, making it a powerful tool for solving complex calculus problems.
Product Rule
The product rule is your friend when differentiating expressions with two functions multiplied together. It tells us that the derivative of two multiplied functions is not just the derivatives multiplied, but a combination of both:
For functions \(u\) and \(v\), the product rule states:
\(\frac{d(uv)}{dx} = u'v + uv'\)
In our exercise, for \(xy+1\), \(xy\) is the product of two separate functions \(x\) and \(y\). So to find the derivative \(\frac{d(xy)}{dx}\), we apply the rule:
  • First term: \(y \frac{d(x)}{dx} = y\)
  • Second term: \(+ x \frac{d(y)}{dx}\)
This helps differentiate mixed terms efficiently, capturing how both variables contribute.
Implicit Function Theorem
Implicit differentiation thrives with equations not easily separable into \(y = f(x)\) form. The implicit function theorem allows us to find derivatives even when functions are tangled together. Rather than solving for \(y\) explicitly, differentiate both sides as they are.

In the given exercise, the equation \((x y + 1)^3 = x - y^2 + 8\) involves both \(x\) and \(y\), requiring differentiation of every part. The implicit function theorem assures us that if the equation defines \(y\) implicitly as a function of \(x\), we can still find \( rac{dy}{dx}\) using implicit differentiation.
This approach taps into the power of calculus even when exact expressions are messy and intertwined.
Differential Calculus
Differential calculus is the study of how things change. By focusing on rates of change, it allows us to calculate slopes and rates in all sorts of functions.
In the world of implicit differentiation, differential calculus provides us with the tools to analyze complex relationships between variables and solve for unknown rates of change easily.
Through techniques like the chain rule and product rule, we're equipped to tackle intricate problems where direct methods aren't applicable. This area of calculus forms the backbone for understanding variable relationships deeply, while also paving the way for future mathematical exploration.

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Most popular questions from this chapter

Vibrations of a spring Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. Use equation (4) to answer the following questions. a. Find the second derivative \(\frac{d^{2} y}{d t^{2}}\) b. Verify that \(\frac{d^{2} y}{d t^{2}}=-\frac{k}{m} y\)

A study conducted at the University of New Mexico found that the mass \(m(t)\) (in grams) of a juvenile desert tortoise \(t\) days after a switch to a particular diet is described by the function \(m(t)=m_{0} e^{0.004 t},\) where \(m_{0}\) is the mass of the tortoise at the time of the diet switch. If \(m_{0}=64\) evaluate \(m^{\prime}(65)\) and interpret the meaning of this result. (Source: Physiological and Biochemical Zoology, 85,1,2012 )

\- Tangency question It is easily verified that the graphs of \(y=1.1^{x}\) and \(y=x\) have two points of intersection, and the graphs of \(y=2^{x}\) and \(y=x\) have no point of intersection. It follows that for some real number \(1.1 < p < 2,\) the graphs of \(y=p^{x}\) and \(y=x\) have exactly one point of intersection. Using analytical and/or graphical methods, determine \(p\) and the coordinates of the single point of intersection.

Suppose \(f\) is differentiable on [-2,2] with \(f^{\prime}(0)=3\) and \(f^{\prime}(1)=5 .\) Let \(g(x)=f(\sin x)\) Evaluate the following expressions. a. \(g^{\prime}(0)\) b. \(g^{\prime}\left(\frac{\pi}{2}\right)\) c. \(g^{\prime}(\pi)\)

Suppose an object of mass \(m\) is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position \(y=0\) when the mass hangs at rest. Suppose you push the mass to a position \(y_{0}\) units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system , the position y of the mass after t seconds is $$y=y_{0} \cos (t \sqrt{\frac{k}{m}})(4)$$ where \(k>0\) is a constant measuring the stiffness of the spring (the larger the value of \(k\), the stiffer the spring) and \(y\) is positive in the upward direction. A damped oscillator The displacement of a mass on a spring suspended from the ceiling is given by \(y=10 e^{-t / 2} \cos \frac{\pi t}{8}\) a. Graph the displacement function. b. Compute and graph the velocity of the mass, \(v(t)=y^{\prime}(t)\) c. Verify that the velocity is zero when the mass reaches the high and low points of its oscillation.
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