- 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:
\(\boxed{ \ (f \cdot g)' = f' \cdot g + f \cdot g' \ }\)
or in differentials notation:
\(\boxed{ \ d(uv)' = udv + vdu \ }\)
Given:
- \(\boxed{ \ u = \theta \ }\)
- \(\boxed{ \ v = sin \ \theta \ }\)
Questions:
\(\boxed{ \ H'(\theta) \ and \ H" (\theta) \ }\)
The Process:
Step-1: determine the first derivative
Remember the following because we will use it:
- \(\boxed{ \ y = sin \ x \rightarrow y' = cos \ x \ }\)
- \(\boxed{ \ y = cos \ x \rightarrow y' = - sin \ x \ }\)
\(\boxed{ \ u = \theta \rightarrow u' = du = 1 \ }\)
\(\boxed{ \ v = sin \ \theta \rightarrow v' = dv = cos \theta \ }\)
\(\boxed{ \ H'(\theta) = udv + vdu \ }\)
\(\boxed{ \ H'(\theta) = (\theta)(cos \theta) + (sin \theta)(1) \ }\)
Thus, we get the first derivative \(\boxed{\boxed{ \ H'(\theta) = \theta \ cos \theta + sin \theta \ }}\)
Step-2: determine the second derivative
From the first derivative \(\boxed{ \ H'(\theta) = \theta \ cos \theta) + sin \theta \ }\) we specify the second derivative.
- Part-1 is \(\boxed{ \ f = \theta \ cos \theta \ }\)
- Part-2 is \(\boxed{ \ g = sin \theta \ }\)
\(\boxed{ \ f' = (\theta)(- sin \ \theta) + (cos \ \theta)(1) \ } \rightarrow \boxed{ \ f' = - \theta sin \ \theta + cos \ \theta \ }\)
\(\boxed{ \ g' = cos \ \theta \ }\)
Let's solve for the second derivative of H.
\(\boxed{ \ H"(\theta) = (- \theta sin \ \theta + cos \ \theta) + (cos \theta) \ }\)
Thus we get the second derivative \(\boxed{\boxed{ \ H"(\theta) = 2cos \theta - \theta sin \ \theta \ }}\)
Learn more
- 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
- The inverse of a function https://brainly.com/question/3225044
- The composite function https://brainly.com/question/1691598
Keywords: the answer, the formula, the product rule, first, the second derivative, differential notation
