Find the derivative of the function.y=sinθ2+cosθy=\frac{\sin\theta}{2+\cos\theta}y=2+cosθsinθ

2 cos ⁡ θ + 1 ( 2 + cos ⁡ θ ) 2 \frac{2\cos\theta+1}{\left(2+\cos\theta\right)^2} ( 2 + c o s θ ) 2 2 c o s θ + 1

Find the derivative of the function. y = sin θ 2 + cos θ y=\frac{\sin\theta}{2+\cos\theta} y = 2 + c o s θ s i n θ

this problem, we're asked to find the derivative of the function y equals sine of theta over two plus cosine theta. So here we have two functions of being divided. We also are dealing with some trig functions, but we do need to use the quotient rule here in order to find this derivative. So remember that our quotient rule tells us low d high minus high d low over the square of what's below. So we're gonna use that in order to take our derivative here. So we're looking for y prime and applying that quotient rule, low d high. I'm going to take that bottom function, my low function, two plus cosine of theta times the derivative of that top function. Now, the derivative of sine theta is just cosine theta. So that is that first term in my numerator. Then I am subtracting my top function, high function, sine of theta d low, so the derivative of that bottom function, two plus cosine theta. Now, the derivative of this bottom function, two plus cosine theta, since I have a constant, the derivative of a constant is zero. I don't have to worry about it. So the derivative of cosine theta is gonna be negative sine theta. So this negative sine theta is the derivative of that bottom function, two plus cosine theta. Now from here, we're dividing this by the square of what's below, which is that two plus cosine theta squared. So from here, we can do some simplifying by actually just multiplying out these two terms to see what we can simplify from there. Now, multiplying that, that will give me two cosine theta plus cosine squared theta. Then that second term, negative sine theta times negative sine theta will give me a positive sine squared theta. And all of that is being divided by two plus cosine theta squared. Now from here, you might think that we're fully done simplifying, but I do notice a trig identity snuck in there because I have cosine theta plus or cosine squared theta plus sine squared theta, which based on my Pythagorean trig identities is equal to one. So we can simplify this a little bit more, setting this equal to one. This is two cosine theta plus one in the top, replacing that with using my trig identity, all divided by two plus cosine theta squared. Now we are fully simplified. This is our derivative here, two cosine theta plus one over two plus cosine theta squared. Now a couple of things that I do want to mention here. You might not always remember to use your trig identities, but I will say that whenever you have cosine squared theta plus sine squared theta, this is probably your most common trig identity. So you should do your best to try to remember this because it will often help you out and help you fully simplify it like we saw here. Now the other thing that I want to mention here is sometimes our denominator, like here we have two plus cosine theta squared, will allow us to more so cancel some stuff out in the top here. Now here, since it's not abundantly obvious that this is going to allow us to cancel anything, I am leaving it as is in parentheses squared. Now if your instructor ever tells you that they want you to try and see if you can cancel out anything else, you can always expand out that denominator and try and cancel what you can here. But more often, you're going to leave it as is with that two plus cosine theta squared. But remember that you should always consult with your instructor to see how they want you to present your answer. Now thanks so much for watching, and let me know if you have questions.

Find the derivative of the function. y = sin ⁡ θ 2 + cos ⁡ θ y=\frac{\sin\theta}{2+\cos\theta} y = 2 + c o s θ s i n θ

2 cos ⁡ θ + 1 ( 2 + cos ⁡ θ ) 2 \frac{2\cos\theta+1}{\left(2+\cos\theta\right)^2} ( 2 + c o s θ ) 2 2 c o s θ + 1

− cos ⁡ θ sin ⁡ θ -\frac{\cos\theta}{\sin\theta} − s i n θ c o s θ

cos ⁡ θ ( 2 + cos ⁡ θ ) 2 \frac{\cos\theta}{\left(2+\cos\theta\right)^2} ( 2 + c o s θ ) 2 c o s θ

12. Trigonometric Functions

Derivatives of Trig Functions

Business Calculus

12. Trigonometric Functions

Derivatives of Trig Functions

Struggling with Business Calculus?

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Multiple Choice

Find the derivative of the function.
$y=\frac{\sin\theta}{2+\cos\theta}$

$-\frac{\cos\theta}{\sin\theta}$

$\frac{\cos\theta}{\left(2+\cos\theta\right)^2}$

$\frac{2\cos\theta+1}{\left(2+\cos\theta\right)^2}$

$\frac{2\cos\theta+\cos^2\theta-\sin^2\theta}{\left(2+\cos\theta\right)^2}$

Verified step by step guidance

Step 1: Recognize that the given function is a rational function, y = (sin(θ)) / (2 + cos(θ)). To find the derivative, we will use the quotient rule, which states that if y = f(θ)/g(θ), then y' = (f'(θ)g(θ) - f(θ)g'(θ)) / (g(θ))^2.

Step 2: Identify f(θ) = sin(θ) and g(θ) = 2 + cos(θ). Compute the derivatives of f(θ) and g(θ). The derivative of f(θ) = sin(θ) is f'(θ) = cos(θ). The derivative of g(θ) = 2 + cos(θ) is g'(θ) = -sin(θ).

Step 3: Apply the quotient rule. Substitute f(θ), f'(θ), g(θ), and g'(θ) into the formula: y' = (f'(θ)g(θ) - f(θ)g'(θ)) / (g(θ))^2. This becomes y' = (cos(θ)(2 + cos(θ)) - sin(θ)(-sin(θ))) / (2 + cos(θ))^2.

Step 4: Simplify the numerator. Expand the terms: cos(θ)(2 + cos(θ)) = 2cos(θ) + cos^2(θ), and -sin(θ)(-sin(θ)) = sin^2(θ). Combine these to get the numerator: 2cos(θ) + cos^2(θ) + sin^2(θ).

Step 5: Use the Pythagorean identity sin^2(θ) + cos^2(θ) = 1 to simplify the numerator further. Replace sin^2(θ) + cos^2(θ) with 1, resulting in the numerator: 2cos(θ) + 1. The final derivative is y' = (2cos(θ) + 1) / (2 + cos(θ))^2.

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