Respuesta :

Answer:

We have:

  • [tex]H'(\theta)=\sin \theta+\theta \cos \theta[/tex]

and

  • [tex]H''(\theta)=2\cos \theta-\theta\sin \theta[/tex]

Step-by-step explanation:

We are given a function H(θ) by:

[tex]H(\theta)=\theta \sin \theta[/tex]

Now, we are asked to find the first and second derivative of the function H(θ)

Now,

[tex]H'(\theta)=\sin \theta(\dfrac{d}{d\theta} \theta)+\theta(\dfrac{d}{d\theta}\sin \theta)\\\\i.e.\\\\H'(\theta)=\sin \theta+\theta \cos \theta[/tex]

Also, the second derivative is  calculate by:

[tex]H''(\theta)=\dfrac{d}{d\theta}\sin \theta+\cos \theta(\dfrac{d}{d\theta}\theta)+\theta(\dfrac{d}{d\theta}\cos \theta)\\\\i.e.\\\\H''(\theta)=\cos \theta+\cos \theta+\theta(-sin \theta)\\\\i.e.\\\\H''(\theta)=2\cos \theta-\theta\sin \theta[/tex]

BladeRunner212
  • H'(θ) = sin θ + θ cos θ
  • H''(θ) = 2cos θ - θ sin θ

Further explanation

In this problem, we will solve the derivatives of products of two functions. The formula used is the product rules as follows:

[tex]\boxed{ \ (f \cdot g)' = f' \cdot g + f \cdot g' \ }[/tex]

or in differentials notation:

[tex]\boxed{ \ d(uv)' = udv + vdu \ }[/tex]

Given:

  • [tex]\boxed{ \ u = \theta \ }[/tex]
  • [tex]\boxed{ \ v = sin \ \theta \ }[/tex]

Questions:

[tex]\boxed{ \ H'(\theta) \ and \ H" (\theta) \ }[/tex]

The Process:

Step-1: determine the first derivative

Remember the following because we will use it:

  • [tex]\boxed{ \ y = sin \ x \rightarrow y' = cos \ x \ }[/tex]
  • [tex]\boxed{ \ y = cos \ x \rightarrow y' = - sin \ x \ }[/tex]

[tex]\boxed{ \ u = \theta \rightarrow u' = du = 1 \ }[/tex]

[tex]\boxed{ \ v = sin \ \theta \rightarrow v' = dv = cos \theta \ }[/tex]

[tex]\boxed{ \ H'(\theta) = udv + vdu \ }[/tex]

[tex]\boxed{ \ H'(\theta) = (\theta)(cos \theta) + (sin \theta)(1) \ }[/tex]

Thus, we get the first derivative [tex]\boxed{\boxed{ \ H'(\theta) = \theta \ cos \theta + sin \theta \ }}[/tex]

Step-2: determine the second derivative

From the first derivative [tex]\boxed{ \ H'(\theta) = \theta \ cos \theta) + sin \theta \ }[/tex] we specify the second derivative.

  • Part-1 is [tex]\boxed{ \ f = \theta \ cos \theta \ }[/tex]
  • Part-2 is [tex]\boxed{ \ g = sin \theta \ }[/tex]

[tex]\boxed{ \ f' = (\theta)(- sin \ \theta) + (cos \ \theta)(1) \ } \rightarrow \boxed{ \ f' = - \theta sin \ \theta + cos \ \theta \ }[/tex]

[tex]\boxed{ \ g' = cos \ \theta \ }[/tex]

Let's solve for the second derivative of H.

[tex]\boxed{ \ H"(\theta) = (- \theta sin \ \theta + cos \ \theta) + (cos \theta) \ }[/tex]

Thus we get the second derivative [tex]\boxed{\boxed{ \ H"(\theta) = 2cos \theta - \theta sin \ \theta \ }}[/tex]

Learn more

  1. If g(x) = (x + 1)/(x - 2) and h(x) = 4 – x, what is the value of g(h(-3))?  https://brainly.com/question/1900154
  2. The inverse of a function https://brainly.com/question/3225044
  3. The composite function https://brainly.com/question/1691598

Keywords: the answer, the formula, the product rule, first, the second derivative, differential notation

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