Textbook Question
45–50. Tangent lines Carry out the following steps. <IMAGE>
a. Verify that the given point lies on the curve.
x³+y³=2xy; (1, 1)
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.8.20a
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
tan xy = x+y; (0,0)
Verified step by step guidance1
Start by differentiating both sides of the equation with respect to x. The equation is \( \tan(xy) = x + y \).
Apply the chain rule to differentiate \( \tan(xy) \). The derivative of \( \tan(u) \) with respect to u is \( \sec^2(u) \), so the derivative of \( \tan(xy) \) with respect to x is \( \sec^2(xy) \cdot (y + x \frac{dy}{dx}) \).
Differentiate the right side of the equation. The derivative of \( x \) with respect to x is 1, and the derivative of \( y \) with respect to x is \( \frac{dy}{dx} \).
Set the derivatives equal to each other: \( \sec^2(xy) \cdot (y + x \frac{dy}{dx}) = 1 + \frac{dy}{dx} \).
Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. Substitute \( x = 0 \) and \( y = 0 \) into the equation to find the specific value of \( \frac{dy}{dx} \) at the point (0,0).
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. Instead of solving for one variable in terms of the other, we differentiate both sides of the equation with respect to the independent variable, applying the chain rule as necessary. This method is particularly useful for equations involving multiple variables or when isolating one variable is complex.
Recommended video:
05:14
Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. It states that if a function y is composed of another function u, then the derivative of y with respect to x can be found by multiplying the derivative of y with respect to u by the derivative of u with respect to x. This rule is essential in implicit differentiation, as it helps to differentiate terms involving products of variables.
Recommended video:
05:02Intro to the Chain Rule
Evaluating at a Point
Evaluating a derivative at a specific point involves substituting the coordinates of that point into the derivative expression. In the context of the given problem, after finding dy/dx using implicit differentiation, we substitute the point (0,0) to determine the slope of the tangent line at that point. This step is crucial for understanding the behavior of the function at specific values and for applications such as finding tangent lines.
Recommended video:
04:50Critical Points
Related Practice
Textbook Question
Surface area of a cone The lateral surface area of a cone of radius r and height h (the surface area excluding the base) is A = πr√r²+h².
a. Find dr/dh for a cone with a lateral surface area of A=1500π.
Textbook Question
60–62. {Use of Tech} Multiple tangent lines Complete the following steps. <IMAGE>
a. Find equations of all lines tangent to the curve at the given value of x.
4x³ =y²(4−x); x=2 (cissoid of Diocles)
1
views
Textbook Question
{Use of Tech} Tree growth Let b represent the base diameter of a conifer tree and let h represent the height of the tree, where b is measured in centimeters and h is measured in meters. Assume the height is related to the base diameter by the function h = 5.67+0.70b+0.0067b².
a. Graph the height function.
Textbook Question
The line tangent to the curve y=h(x) at x=4 is y = −3x+14. Find an equation of the line tangent to the following curves at x=4.
y = (x²-3x)h(x)
Textbook Question
Witch of Agnesi Let y(x²+4)=8 (see figure). <IMAGE>
a. Use implicit differentiation to find dy/dx.