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Find the derivatives of the functions. $$r=6(\sec \theta-\tan \theta)^{3 / 2}$$

Short Answer

Expert verified
The derivative is \( 9(\sec \theta - \tan \theta)^{1/2}(\tan \theta - \sec \theta) \).

Step by step solution

01

Identify Function Form and Apply Chain Rule

The given function is \( r = 6(\sec \theta - \tan \theta)^{3/2} \). Notice that it is a composition of functions where \( u = \sec \theta - \tan \theta \) and the outer function is \( u^{3/2} \). To differentiate \( r \) with respect to \( \theta \), apply the chain rule: \( \frac{dr}{d\theta} = \frac{dr}{du} \times \frac{du}{d\theta} \).
02

Differentiate Outer Function

Differentiate the outer function \( u^{3/2} \) with respect to \( u \). The derivative is \( \frac{d}{du}(u^{3/2}) = \frac{3}{2}u^{1/2} \). This gives us \( \frac{dr}{du} = \frac{3}{2}(6)(\sec \theta - \tan \theta)^{1/2} \). Hence, \( \frac{dr}{du} = 9(\sec \theta - \tan \theta)^{1/2} \).
03

Differentiate Inner Function

Next, differentiate the inner function \( u(\theta) = \sec \theta - \tan \theta \) with respect to \( \theta \). Recall \( \sec \theta = \frac{1}{\cos \theta} \) which gives \( \frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta \). Also, \( \frac{d}{d\theta}(\tan \theta) = \sec^2 \theta \). Thus, \( \frac{du}{d\theta} = \sec \theta \tan \theta - \sec^2 \theta \).
04

Multiply Derivatives from the Chain Rule

Multiply the derivatives from Step 2 and Step 3 to apply the chain rule: \( \frac{dr}{d\theta} = 9(\sec \theta - \tan \theta)^{1/2} \times (\sec \theta \tan \theta - \sec^2 \theta) \).
05

Simplify the Expression

Simplify the expression from Step 4: \( 9(\sec \theta - \tan \theta)^{1/2}(\sec \theta - \tan \theta \tan \theta) \). Simplification might vary depending on context but outputs the derivative in a simplified form. The final expression is \( 9(\sec \theta - \tan \theta)^{1/2}(\tan \theta - \sec \theta) \).

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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 fundamental technique in calculus for finding the derivative of composite functions. When a function is composed of two or more functions, the chain rule helps in breaking down the differentiation process into manageable steps. The basic idea is to differentiate the outer function and multiply it by the derivative of the inner function.

To apply the chain rule, consider a composite function such as \( f(g(x)) \). The derivative with respect to \( x \) is given by \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \). This means you first take the derivative of the outer function \( f \) with respect to the inner function \( g(x) \), and then multiply it by the derivative of the inner function \( g \) with respect to \( x \).

Here's a quick example: If \( y = (x^2 + 1)^3 \), let \( u = x^2 + 1 \). Then \( y = u^3 \), so \( \frac{dy}{dx} = 3u^2 \cdot \frac{du}{dx} = 3(x^2 + 1)^2 \cdot 2x = 6x(x^2 + 1)^2 \). The chain rule is powerful for any function that "nests" other functions within it.
Trigonometric Functions
Trigonometric functions such as \( \sin \theta, \cos \theta, \tan \theta, \sec \theta \) and others are the cornerstones of trigonometry. They relate the angles of a triangle to the lengths of its sides. In calculus, these functions are often differentiated to find rates of change concerning an angle. Differentiation of these functions utilizes specific derivative rules, such as:
  • \( \frac{d}{d\theta}(\sin \theta) = \cos \theta \)
  • \( \frac{d}{d\theta}(\cos \theta) = -\sin \theta \)
  • \( \frac{d}{d\theta}(\tan \theta) = \sec^2 \theta \)
  • \( \frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta \)
Trigonometric identities also play a crucial role when simplifying derivatives. For instance, \( \sec \theta \) is 1 over \( \cos \theta \), and its derivative results in \( \sec \theta \tan \theta \) due to the quotient rule and trigonometric identity \( \sec^2 \theta = \tan^2 \theta + 1 \).

Understanding these derivative rules and identities aids in analyzing curves and motion in spaces where angles are significant.
Derivative of Composite Functions
Differentiating composite functions involves applying the chain rule, as discussed earlier. A composite function is one where a function is nested inside another, like \( h(x) = f(g(x)) \). When such functions involve trigonometric expressions, extra care must be taken to differentiate them correctly.

Consider a function \( r = 6(\sec \theta - \tan \theta)^{3/2} \). This is a composite function where the inner function is \( u = \sec \theta - \tan \theta \), and the outer function is \( u^{3/2} \). The derivative is determined by applying the chain rule: first, differentiate the outer function, \( u^{3/2} \), yielding \( \frac{3}{2}u^{1/2} \); then, differentiate the inner function \( \sec \theta - \tan \theta \), resulting in \( \sec \theta \tan \theta - \sec^2 \theta \).

Finally, multiply these derivatives to get the full derivative: \( 9(\sec \theta - \tan \theta)^{1/2}(\sec \theta \tan \theta - \sec^2 \theta) \). Each step requires a clear understanding of basic derivative rules and the ability to manipulate and simplify trigonometric expressions.

Utilizing both trigonometric identities and algebraic manipulation, the derivative of composite functions becomes easier to handle and understand.

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Most popular questions from this chapter

Find \(d y / d t\) $$y=\left(1+\tan ^{4}\left(\frac{t}{12}\right)\right)^{3}$$

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x y^{3}+\tan (x+y)=1, \quad P\left(\frac{\pi}{4}, 0\right)$$

A highway patrol plane flies 3 km above a level, straight road at a steady \(120 \mathrm{km} / \mathrm{h}\). The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is \(5 \mathrm{km},\) the line-of-sight distance is decreasing at the rate of \(160 \mathrm{km} / \mathrm{h}\). Find the car's speed along the highway.

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x^{5}+y^{3} x+y x^{2}+y^{4}=4, \quad P(1,1)$$

Suppose that the functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=0\) and \(x=1\) $$\begin{array}{ccccc} \hline x & f(x) & g(x) & f^{\prime}(x) & g^{\prime}(x) \\ \hline 0 & 1 & 1 & 5 & 1 / 3 \\ 1 & 3 & -4 & -1 / 3 & -8 / 3 \\ \hline \end{array}$$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x\) a. \(5 f(x)-g(x), \quad x=1\) b. \(f(x) g^{3}(x), \quad x=0\) c. \(\frac{f(x)}{g(x)+1}, \quad x=1\) d. \(f(g(x)), \quad x=0\) e. \(g(f(x)), \quad x=0\) f. \(\left(x^{11}+f(x)\right)^{-2}, \quad x=1\) g. \(f(x+g(x)), \quad x=0\)

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