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Find the derivative. Simplify where possible. $$y=\arctan (\tanh x)$$

Short Answer

Expert verified
The derivative is 1.

Step by step solution

01

Understand the Derivative Chain Rule

The function we have is a composition of two functions: the arctangent function and the hyperbolic tangent function. To find the derivative, we will use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
02

Differentiate the Outer Function

The outer function is the inverse tangent, or arctan function. The derivative of the arctan function, with respect to its argument, is \( \frac{1}{1 + u^2} \), where \( u = \tanh(x) \). Thus, the derivative of the outer function with respect to the inner function is \( \frac{1}{1 + (\tanh(x))^2} \).
03

Differentiate the Inner Function

The inner function is the hyperbolic tangent function \( \tanh(x) \). The derivative of \( \tanh(x) \) is \( \text{sech}^2(x) \). This results from differentiating the identity \( \tanh(x) = \frac{\sinh(x)}{\cosh(x)} \), where the derivative of \( \sinh(x) \) is \( \cosh(x) \) and the derivative of \( \cosh(x) \) is \( \sinh(x) \).
04

Apply the Chain Rule

Now, apply the chain rule by multiplying the derivatives found in Step 2 and Step 3. The derivative of \( y = \arctan(\tanh(x)) \) becomes \( \frac{1}{1 + (\tanh(x))^2} \cdot \text{sech}^2(x) \).
05

Simplify the Expression

To simplify, recall the identity \( 1 + \tanh^2(x) = \text{sech}^2(x) \). Substitute this identity into the expression: \( \frac{1}{\text{sech}^2(x)} \cdot \text{sech}^2(x) = 1 \). Therefore, the simplified derivative is \( 1 \).

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Key Concepts

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

Chain Rule
When dealing with derivatives, the chain rule is a crucial tool, especially when a function is composed of other functions. The chain rule helps us find the derivative of a composite function by differentiating each part separately. In essence, it states that the derivative of a composite function \( f(g(x)) \) is given by \( f'(g(x)) \cdot g'(x) \).

1. The **outer function**: This is the function that is applied last in the composition. We differentiate this function while treating the inner function as its variable.
2. The **inner function**: This is the function inside the outer function. We differentiate this next as if it were independent.

This approach allows us to tackle complex derivatives step-by-step. In problems involving inverse and hyperbolic functions, mastering the chain rule makes differentiation more manageable.
Inverse Trigonometric Functions
Inverse trigonometric functions, like the arctangent \( \arctan(x) \), help us find angles when given a ratio of sides in a right-angled triangle. In calculus, these functions have specific derivatives that are important to remember:

- For \( \arctan(u) \), the derivative is \( \frac{1}{1 + u^2} \).

This formula provides the rate at which the angle changes as the input \( u \) changes. In the example provided, \( u = \tanh(x) \). Therefore, knowing this derivative is a key step in applying the chain rule when combined with other functions, such as hyperbolic. Understanding the behavior of inverse trigonometric functions is vital for calculus students as they often appear in complex differentiation and integration problems.
Hyperbolic Functions
Hyperbolic functions resemble trigonometric functions but are based on hyperbolas instead of circles. An important hyperbolic function is \( \tanh(x) \), defined as \( \frac{\sinh(x)}{\cosh(x)} \).

1. **Tangent Hyperbolic, \( \tanh(x) \):** Represents the ratio between the hyperbolic sine and cosine.
2. **Derivative of \( \tanh(x) \):** This is \( \text{sech}^2(x) \), where \( \text{sech}(x) \) is the hyperbolic secant function, calculated from \( 1/\cosh(x) \).

The identity \( 1 + \tanh^2(x) = \text{sech}^2(x) \) plays a crucial role in simplifying derivatives involving hyperbolic functions. These functions offer unique properties in calculus, making them indispensable for certain integrals and derivatives. Understanding their derivatives is necessary for complex calculus problems and improves your overall mathematical toolkit.

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Most popular questions from this chapter

A plane flies horizontally at an altitude of 5 \(\mathrm{km}\) and passes directly over a tracking telescope on the ground. When the angle of elevation is \(\pi / 3,\) this angle is decreasing at a rate of \(\pi / 6 \mathrm{rad} / \mathrm{min}\) . How fast is the plane traveling at that time?

The gas law for an ideal gas at absolute temperature T (in kelvins), pressure \(P\) (in atmospheres), and volume \(V\) ( in liters) is \(P V=n R T\) , where \(n\) is the number of moles of the gas and \(R=0.0821\) is the gas constant. Suppose that, at a certain instant, \(\mathrm{P}=8.0\) atm and is increasing at a rate of 0.10 atm/min and \(\mathrm{V}=10 \mathrm{L}\) and is decreasing at a rate of 0.15 \(\mathrm{L} / \mathrm{min}\) . Find the rate of change of \(\mathrm{T}\) with respect to time at that instant if \(\mathrm{n}=10 \mathrm{mol} .\)

\(37-48\) Use logarithmic differentiation to find the derivative of the function. $$y=(\tan x)^{1 / x}$$

(a) If \(\$ 3000\) is invested at 5\(\%\) interest, find the value of the investment at the end of 5 years if the interest is compounded ( i ) annually, (ii) semiannually, (iil) monthly, (iv) weekly, (v) daily, and (vi) continuously. (b) If \(\mathrm{A}(\mathrm{t})\) is the amount of the investment at time t for the case of continuous compounding, write a differential equation and an initial condition satisfied by \(\mathrm{A}(\mathrm{t})\) .

(a) If \(n\) is a positive integer, prove that $$\frac { d } { d x } \left( \sin ^ { n } x \cos n x \right) = n \sin ^ { n - 1 } x \cos ( n + 1 ) x$$ (b) Find a formula for the derivative of \(y = \cos ^ { n } x \cos n x\) that is similar to the one in part (a).

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