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Chapter 3: Problem 5

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\) b. Find the slope of the curve at the given point. $$x^{4}+y^{4}=2 ;(1,-1)$$

Short Answer

Expert verified
Using implicit differentiation, we found that the derivative of the given equation, \(x^4+y^4=2\), is \(\frac{dy}{dx} = \frac{-x^3}{y^3}\). By plugging the coordinates of the given point (1, -1) into the expression for \(\frac{dy}{dx}\), we found that the slope of the curve at this point is 1.

Step by step solution

01

Apply Implicit Differentiation

To find the derivative \(\frac{dy}{dx}\), we'll apply implicit differentiation to the given equation, \(x^4+y^4=2\). Differentiate both sides of the equation with respect to x: $$\frac{d}{dx}(x^4+y^4) = \frac{d}{dx}(2)$$ Now, differentiate \(x^4\) and \(y^4\) separately: $$\frac{d}{dx}(x^4) + \frac{d}{dx}(y^4) = 0$$ The derivative of \(x^4\) with respect to x is \(4x^3\). To find the derivative of \(y^4\) with respect to x, we apply the chain rule since we need to differentiate y with respect to x: $$4x^3 + 4y^3\frac{dy}{dx} = 0$$ Now we can solve for \(\frac{dy}{dx}\).
02

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

To solve for \(\frac{dy}{dx}\), we first isolate it on one side of the equation: $$4y^3\frac{dy}{dx} = -4x^3$$ Now, divide both sides by \(4y^3\): $$\frac{dy}{dx} = \frac{-4x^3}{4y^3}$$ Simplifying gives us the derivative: $$\frac{dy}{dx} = \frac{-x^3}{y^3}$$
03

Find the slope at the given point

Now, we'll plug the coordinates of the given point (1, -1) into the expression for \(\frac{dy}{dx}\) to compute the slope of the curve at that point: $$\frac{dy}{dx} = \frac{-(1)^3}{(-1)^3}$$ Which simplifies to: $$\frac{dy}{dx} = \frac{-1}{-1}$$ Thus, the slope of the curve at the given point (1, -1) is: $$\frac{dy}{dx} = 1$$

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Key Concepts

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

Chain Rule
The chain rule is an essential tool when dealing with implicit differentiation. It helps us find the derivative of a composite function. In simple terms, when you have a function inside another function, you need to apply the chain rule to differentiate it with respect to an independent variable. In our specific case, the equation given is \(x^4 + y^4 = 2\). Here, \(y\) is a function of \(x\), even if it's not explicitly stated. When differentiating \(y^4\) with respect to \(x\), you apply the chain rule by multiplying the derivative of \(y^4\) with respect to \(y\), which is \(4y^3\), by \(\frac{dy}{dx}\). This application of the chain rule is necessary because \(y\) is dependent on \(x\). Without the chain rule, handling implicit dependencies would be cumbersome.
Curve Slope
The slope of a curve at a given point tells us how steep the graph is at that instant. It is found using the derivative and tells us the rate of change of the function with respect to the independent variable. In the context of the exercise, once the derivative \(\frac{dy}{dx}\) was determined as \(\frac{-x^3}{y^3}\), we used the given point
  • (1, -1)
for substitution. By substituting \(x = 1\) and \(y = -1\) into our derivative, we simplify to find the slope. Calculating the slope at this specific point helps to understand how the curve behaves locally at that spot on the graph. In this case, upon substituting, the slope came out to be 1, meaning at the point \((1,-1)\), the curve is rising at a 45-degree angle, which is quite intuitive when visualizing the graph's behavior at that point.
Derivatives
Derivatives are the backbone of calculus, used to measure how a function changes as its inputs change. In this exercise, the derivative \(\frac{dy}{dx}\) was determined through implicit differentiation due to the form \(x^4 + y^4 = 2\). Finding derivatives involves differentiating each term individually and applying rules like the chain rule where necessary. This derivative expression essentially provides the slope of the tangent line to the curve at any given point \((x, y)\). Derivatives play a crucial role in evaluating how functions behave and change, and are also extensively applied across various fields such as physics, engineering, and economics. Understanding implicit differentiation and derivative computation through these basic rules enables solving complex real-world problems involving changing quantities.

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Most popular questions from this chapter

Earth's atmospheric pressure decreases with altitude from a sea level pressure of 1000 millibars (a unit of pressure used by meteorologists). Letting \(z\) be the height above Earth's surface (sea level) in \(\mathrm{km}\), the atmospheric pressure is modeled by \(p(z)=1000 e^{-z / 10}.\) a. Compute the pressure at the summit of Mt. Everest which has an elevation of roughly \(10 \mathrm{km}\). Compare the pressure on Mt. Everest to the pressure at sea level. b. Compute the average change in pressure in the first \(5 \mathrm{km}\) above Earth's surface. c. Compute the rate of change of the pressure at an elevation of \(5 \mathrm{km}\). d. Does \(p^{\prime}(z)\) increase or decrease with \(z\) ? Explain. e. What is the meaning of \(\lim _{z \rightarrow \infty} p(z)=0 ?\)

a. Differentiate both sides of the identity \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\) to prove that \(\sin 2 t=2 \sin t \cos t\). b. Verify that you obtain the same identity for sin \(2 t\) as in part (a) if you differentiate the identity \(\cos 2 t=2 \cos ^{2} t-1\). c. Differentiate both sides of the identity \(\sin 2 t=2 \sin t \cos t\) to prove that \(\cos 2 t=\cos ^{2} t-\sin ^{2} t\).

Find \(\frac{d y}{d x},\) where \(\left(x^{2}+y^{2}\right)\left(x^{2}+y^{2}+x\right)=8 x y^{2}\)

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{x \rightarrow e} \frac{\ln x-1}{x-e}$$

Proof by induction: derivative of \(e^{k x}\) for positive integers \(k\) Proof by induction is a method in which one begins by showing that a statement, which involves positive integers, is true for a particular value (usually \(k=1\) ). In the second step, the statement is assumed to be true for \(k=n\), and the statement is proved for \(k=n+1,\) which concludes the proof. a. Show that \(\frac{d}{d x}\left(e^{k x}\right)=k e^{k x},\) for \(k=1\) b. Assume the rule is true for \(k=n\) (that is, assume \(\left.\frac{d}{d x}\left(e^{n x}\right)=n e^{n x}\right),\) and show this assumption implies that the rule is true for \(k=n+1\). (Hint: Write \(e^{(n+1) x}\) as the product of two functions and use the Product Rule.)
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