Find the points on the curve y=(cosx) /(2+sinx) at which the tangent is horizontal.

Name: Problem 34, Chapter 2: Calculus
Uploaded: 2021-04-07T19:11:01-08:00
Duration: 11 min 5 s
Channel: Adam Schade
Description: Find the points on the curve y=(cosx) /(2+sinx) at which the tangent is horizontal.

Find the points on the curve y=(cosx) /(2+sinx) at which the...

Key Concepts

Quotient Rule

The quotient rule is a method used to differentiate functions that are expressed as the ratio of two differentiable functions. It states that the derivative of a function h(x) = f(x)/g(x) is given by (g(x)f'(x) - f(x)g'(x))/(g(x))^2. This rule is crucial when dealing with rational functions and helps in systematically obtaining the derivative.

Trigonometric Functions

Trigonometric functions such as sine and cosine play a significant role in many calculus problems because of their periodic properties and well-known derivatives. Proficiency in differentiating these functions and applying trigonometric identities is essential for solving problems involving curves defined by trigonometric expressions.

Differentiation

Differentiation is the process of finding the derivative, which represents the rate at which a function changes. In problems involving curves, finding the derivative is essential to determine the slope of the tangent line at any given point, which is a fundamental concept in calculus.

Horizontal Tangents

A horizontal tangent occurs at points on a curve where the slope of the tangent line is zero. This is determined by setting the derivative equal to zero and solving for the corresponding values, which identifies the critical points where the function's rate of change momentarily is flat.

Transcript

00:02 Okay, so the question we're going to be working on this time is find the point on the curve.

00:10 Y equals cosine of x over 2 plus sine of x where the tangent is horizontal.

00:21 So this one's a bit more tricky, a bit more involved.

00:25 So the function, y is equal to cosine of x all over 2 plus.

00:41 Sign of x.

00:44 It wants us to find where the tangent line is horizontal, which is basically the same way as saying that it wants to find where the derivative of the function is equal to zero, which means we have to derive this function.

01:00 So, of course, this is a division.

01:07 This has a division in it, so that means we have to use the division rule, which is always fun.

01:16 My little memorization thing to always remember the division rule is low d high minus high d low all over low squared.

01:28 The little thing my teacher taught me.

01:32 So let's do that.

01:34 Low d high, that would be 2 plus sine of x times the derivative cosine is negative sine of x.

01:48 Minus high d low so that'd be cosine of x times the derivative of the bottom which is just that just be cosine of x because the two disappears because it's a constant okay all over low squared which would be two plus sine of x squared so we can let's simplify a little bit but let's not go to overboard um again prime be equal to let's leave the first part as it is instead let's move the minus sign out so that'd be minus two plus sine of x times sign of x minus we can combine the cosines though to squared of x all over 2 plus sine of x squared so now we have to find the zeros of this function the values of x that make y prime equal to zero and big thing you have to look at here is we have a denominator we have to make sure we don't include any values where the denominator is also equal to zero.

04:02 Important thing to remember.

04:04 So looking at this.

04:10 Okay, so i actually had to pause there to sort of figure out what we're going to do.

04:15 So obviously, this is a bit too complicated in order to just go straight and find the zeros of the function.

04:26 So you have to simplify it a little bit.

04:29 So the first thing i'm going to do is both, multiply the sine x out to the other sine x and the two, as well as take out the negative signs and move it up front...

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