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Chapter 2: Problem 54

Find the derivative of the function. $$ g(t)=\tan (\cos 2 t) $$

Short Answer

Expert verified
The derivative of the function \(g(t) = \tan(\cos 2t)\) is: \[\frac{dg}{dt} = -2\sin(2t)\sec^2(\cos 2t)\]

Step by step solution

01

Identify the chain rule formula

The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. In other words, if we have a function \(y = f(g(x))\), then its derivative with respect to x is given by: \[ \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \]
02

Apply the chain rule

In our case, \(g(t) = \tan(\cos 2t)\). Let's rewrite it as a composition of two functions: \[f(u) = \tan(u) \quad \text{and} \quad u = \cos 2t\] Now we need to find the derivatives of both functions with respect to their respective variables: 1. Derivative of \(f(u)\) with respect to u: \[ \frac{df}{du} = \frac{d}{du}\tan(u) \] 2. Derivative of \(u\) with respect to t: \[ \frac{du}{dt} = \frac{d}{dt}(\cos 2t) \]
03

Find the derivatives of the individual functions

1. The derivative of \(f(u) = \tan(u)\) with respect to u is: \[ \frac{df}{du} = \frac{d}{du}\tan(u) = \sec^2(u) \] 2. The derivative of \(u = \cos 2t\) with respect to t is: \[\frac{du}{dt} = \frac{d}{dt}(\cos 2t) = -2\sin(2t)\]
04

Use chain rule to compute the derivative of the main function

Now using the chain rule to find the derivative of the main function \(g(t) = \tan(\cos 2t)\): \[\frac{dg}{dt} = \frac{df}{du} \cdot \frac{du}{dt} = \sec^2(u) \cdot (-2\sin(2t))\] Since \(u = \cos 2t\), substitute back in the expression: \[\frac{dg}{dt} = \sec^2(\cos 2t) \cdot (-2\sin(2t))\] So, the derivative of the function \(g(t) = \tan(\cos 2t)\) is: \[\frac{dg}{dt} = -2\sin(2t)\sec^2(\cos 2t)\]

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Key Concepts

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

Chain Rule
The chain rule is a powerful tool in calculus for finding the derivative of composite functions. When we have a function of a function, the chain rule helps us work out the overall derivative by focusing on each part. Consider a function like \( y = f(g(x)) \). The chain rule tells us to differentiate the outer function \( f \) and multiply it by the derivative of the inner function \( g \). The formula is:
  • \( \frac{dy}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \)
The beauty of the chain rule makes it easier to handle complex layers in a function. Simply treat the inner function as a temporary variable and differentiate step by step.
Composite Functions
Composite functions are like nesting functions inside one another. For example, in \( g(t) = \tan(\cos 2t) \), there are two distinct functions: \( \tan(u) \) as the outer function and \( \cos 2t \) as the inner function. You break it down as:
  • \( f(u) = \tan(u) \)
  • \( u = \cos 2t \)
Understanding this nesting is essential. It allows you to differentiate complex expressions by dealing with simpler parts. Start by differentiating the function fully within its own context (like \( u = \cos 2t \)), then handle the outer layer with its result (like \( f(u) = \tan(u) \)). This process helps you manage complex derivatives piece by piece.
Trigonometric Derivatives
Derivatives of trigonometric functions are a key part of calculus, helping analyze periodic patterns. Recognizing these derivatives can make solving problems much simpler:
  • \( \frac{d}{du} \tan(u) = \sec^2(u) \)
  • \( \frac{d}{dt} \cos(2t) = -2\sin(2t) \)
Trigonometric derivatives serve as building blocks for understanding various phenomena. In our exercise, \( \tan \) and \( \cos \) provide their own challenges but bring unique methods for handling derivatives. Knowing how to differentiate them enhances problem-solving skills and allows you to apply these techniques to similar problems across different fields.

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Most popular questions from this chapter

Suppose that \(u\) is a differentiable function of \(x\) and \(f(x)=|u| .\) Show that $$ f^{\prime}(x)=\frac{u^{\prime} u}{|u|} \quad u \neq 0 $$ Hint: \(|u|=\sqrt{u^{2}}\).

Orbit of a Satellite An artificial satellite moves around the earth in an elliptic orbit. Its distance \(r\) from the center of the earth is approximated by $$ r=a\left[1-e \cos M-\frac{e^{2}}{2}(\cos 2 M-1)\right] $$ where \(M=(2 \pi / P)\left(t-t_{n}\right) .\) Here, \(t\) is time and \(a, e, P\), and \(t_{n}\) are constants measuring the semimajor axis of the orbit, the eccentricity of the orbit, the period of orbiting, and the time taken by the satellite to pass the perigee, respectively. Find \(d r / d t\), the radial velocity of the satellite.

Evaluate the limit using l'H么pital's Rule if appropriate. $$ \lim _{x \rightarrow \infty}\left(x-\sqrt{x^{2}+1}\right) $$

Find the derivative of the function. $$ f(x)=\sin ^{-1} 3 x $$

Find an equation of the tangent line to the given curve at the indicated point. $$ 2 y^{2}-x^{3}-x^{2}=0 ; \quad(1,1) $$
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