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Find the derivatives of the functions. Assume \(a, b,\) and \(c\) are constants. $$f(\theta)=\theta^{3} \cos \theta$$

Short Answer

Expert verified
The derivative is \(f'(\theta) = 3\theta^{2} \cos \theta - \theta^{3} \sin \theta\).

Step by step solution

01

Identify Function Parts

The function given is a product of two functions: \(u = \theta^{3}\) and \(v = \cos \theta\). To differentiate this function, we will use the product rule, which states that for two functions \(u\) and \(v\), the derivative \(f'(\theta)\) is \(u'v + uv'\).
02

Differentiate the First Part \(u\)

Differentiate \(u = \theta^{3}\) with respect to \(\theta\). Using the power rule, the derivative \(u'\) is given by: \[u' = \frac{d}{d\theta}(\theta^{3}) = 3\theta^{2}\]
03

Differentiate the Second Part \(v\)

Differentiate \(v = \cos \theta\) with respect to \(\theta\). The derivative of \(\cos \theta\) is \(-\sin \theta\). Therefore,\[v' = \frac{d}{d\theta}(\cos \theta) = -\sin \theta\]
04

Apply the Product Rule

Using the product rule formula \( u'v + uv'\), substitute the derivatives and the original functions:\[f'(\theta) = (3\theta^{2})(\cos \theta) + (\theta^{3})(-\sin \theta)\]Simplify the expression by distributing:\[f'(\theta) = 3\theta^{2}\cos \theta - \theta^{3}\sin \theta\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Product Rule
The product rule is a fundamental tool in calculus, helping us to find the derivative of a product of two functions. It's essential whenever you encounter a function that's the product of two simpler functions. The rule states that if you have two differentiable functions, say \( u(x) \) and \( v(x) \), then the derivative of their product \( f(x) = u(x)v(x) \) is given by:
  • \( f'(x) = u'(x)v(x) + u(x)v'(x) \)
This formula might look a bit daunting at first, but it's just an application of both functions' derivatives. Here's a simple breakdown:
  • First, find the derivative of the first function, keeping the second function the same.
  • Then, find the derivative of the second function, while keeping the first function the same.
  • Multiply and add these results.
Using the product rule is like performing a dance: order matters. It ensures that each function's change is accounted for when combined with its partner.
Next time you see a couple of functions multiplying each other, don't fret – use the product rule to separate and conquer.
Power Rule
The power rule is one of the simplest and most widely used rules in differentiation. It's applied when you're working with a power function, such as \( \theta^n \). The rule allows us to easily take derivatives of these functions. The power rule states that if \( f(x) = x^n \), then the derivative \( f'(x) \) is:
  • \( f'(x) = nx^{n-1} \)
In plain terms, to differentiate \( x^n \), move the power \( n \) in front of \( x \) and decrease the power by 1. The power rule is effortlessly handy:
  • No matter the starting exponent, the process remains the same.
  • If \( n = 3 \), like in \( \theta^3 \), simply multiply by 3 and drop the power to \( \theta^{2} \).
These predictable steps turn the chaos of unfamiliar exponents into ordered simplicity. This rule is a bedrock for handling polynomial functions, making it just as crucial as your morning coffee.
Trigonometric Derivatives
Trigonometric derivatives are the building blocks when differentiating functions containing trigonometric terms such as sine, cosine, or tangent. Each fundamental trigonometric function has a simple derivative:
  • The derivative of \( \sin x \) is \( \cos x \).
  • The derivative of \( \cos x \) is \( -\sin x \).
  • The derivative of \( \tan x \) is \( \sec^2 x \).
Knowing these derivatives allows us to navigate more complex expressions involving trigonometric functions. For example, the solution above required the derivative of \( \cos \theta \), which is \( -\sin \theta \).
When differentiating a trigonometric function, remember:
  • These derivatives stem from their respective function’s change in behavior.
  • Each trigonometric rule has a negative or a square involved, which must be remembered.
Grasp these basic derivatives, and you'll increase your proficiency in untangling equations with both simplicity and confidence.

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Most popular questions from this chapter

Find the derivative. It may be to your advantage to simplify before differentiating. Assume \(a, b, c\) and \(k\) are constants. $$T(u)=\arctan \left(\frac{u}{1+u}\right)$$

A rubber duck bounces up and down in a pool after a stone is dropped into the water. The height of the duck, in inches, above the equilibrium position of the water is given as a function of time \(t,\) in seconds, by $$d(t)=e^{-t}(\cos t+\sin t)$$ (a) Find and interpret the practical meaning of the derivative \(d^{\prime}(t)\) (b) Determine when \(d^{\prime}(t)=0\) for \(t \geq 0 .\) What can you say about the duck when \(d^{\prime}(t)=0 ?\) (c) Determine \(\lim _{t \rightarrow \infty} d(t)\) and explain why this limit makes sense in practical terms.

Normal human body temperature fluctuates with a rhythm tied to our sleep cycle. \(^{14}\) If \(H(t)\) is body temperature in degrees Celsius at time \(t\) in hours since 9 am, then \(H(t)\) may be modeled by $$H(t)=36.8+0.6 \sin \left(\frac{\pi}{12} t\right)$$ (a) Calculate \(H^{\prime}(t)\) and give units. (b) Calculate \(H^{\prime}(4)\) and \(H^{\prime}(12),\) then interpret the meaning of your answers in everyday terms.

(a) Show that \(1-x\) is the local linearization of \(\frac{1}{1+x}\) near \(x=0\) (b) From your answer to part (a), show that near \(x=0\) $$ \frac{1}{1+x^{2}} \approx 1-x^{2} $$ (c) Without differentiating, what do you think the derivative of \(\frac{1}{1+x^{2}}\) is at \(x=0 ?\)

Find and interpret the value of the expression in practical terms. Let \(\bar{C}(t)\) be the concentration of carbon dioxide in parts per million (ppm) in the air as a function of time, \(t,\) in months since December 1,2005: 15 $$C(t)=3.5 \sin \left(\frac{\pi t}{6}\right)+381+\frac{t}{6}$$ $$C^{\prime}(30)$$

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