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Evaluate the derivatives of the following functions. $$f(x)=1 / \tan ^{-1}\left(x^{2}+4\right)$$

Short Answer

Expert verified
Answer: The derivative of the function is $$f'(x)=\frac{2x}{\left[\tan^{-1}(x^2+4)\right] \left[1+(x^2+4)^2\right]}$$.

Step by step solution

01

Identify the functions

In this exercise, we have the function $$f(x)=\frac{1}{\tan^{-1}(x^2+4)}$$. We can rewrite this as $$f(x)=\left[ \tan^{-1}(x^2+4) \right]^{-1} .$$ Now we can identify the two functions: $$g(x) = x^2+4,$$ $$h(x) = \tan^{-1}(x).$$ Our function $$f(x)= h(g(x))^(-1).$$
02

Find g'(x) and h'(x)

Next, we find the derivatives of g(x) and h(x): $$g'(x) = \frac{d(x^2+4)}{dx}=2x,$$ $$h'(x) = \frac{d(\tan^{-1}(x))}{dx}=\frac{1}{1+x^2}. $$ Now we have all the information we need to apply the chain rule.
03

Apply the chain rule

Using the chain rule, we find the derivative of f(x): $$f'(x) = \left[h(g(x))\right]^{-1} * h'(g(x)) * g'(x)$$ $$f'(x) = \left[\tan^{-1}(x^2+4)\right]^{-1} * \frac{1}{1+(x^2+4)^2} * 2x$$
04

Simplify the result

Now we can simplify the expression: $$f'(x) = \frac{2x}{\left[\tan^{-1}(x^2+4)\right] \left[1+(x^2+4)^2\right]}$$ The derivative of the function $$f(x)=\frac{1}{\tan^{-1}(x^2+4)}$$ is $$f'(x)=\frac{2x}{\left[\tan^{-1}(x^2+4)\right] \left[1+(x^2+4)^2\right]}$$.

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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 used to find the derivative of composite functions. A composite function is simply one function nested inside another. When evaluating the derivative of such functions, the chain rule allows us to handle the layers separately.

Think of it like peeling an onion—each layer is derived individually before combining them all. In mathematical terms, if you have functions \(f(x) = h(g(x))\), the chain rule states that the derivative \(f'(x)\) is found by multiplying the derivative of the outer function \(h'(x)\), evaluated at \(g(x)\), by the derivative of the inner function \(g'(x)\). The formula looks like this:
  • \(f'(x) = h'(g(x)) \cdot g'(x)\)
By mastering the chain rule, you'll gain the ability to work with more complex function compositions and find derivatives in a systematic way.
Inverse Trigonometric Functions
Inverse trigonometric functions are the functions that reverse the roles of the trigonometric functions. For example, if the function \(y = an(x)\) finds the tangent of \(x\), then the inverse function \(x = an^{-1}(y)\) finds the angle \(x\) where \(y = an(x)\).

Inverse trigonometric functions include \( an^{-1}(x)\), \( ext{sin}^{-1}(x)\), \( ext{cos}^{-1}(x)\), and others. They are used when you need to find the angle that results in a given trigonometric value. However, these functions come with restrictions in their range to ensure they are valid functions and pass the vertical line test. Understanding the derivatives of these functions, such as \( \frac{d}{dx}( an^{-1}(x)) = \frac{1}{1+x^2} \), is crucial for solving calculus problems involving these inverse functions. They often come into play when using the chain rule.
Function Composition
Function composition is the process of combining two functions. If you have a function \(g(x)\) and another function \(h(x)\), composing them means inserting one function into another, resulting in a new function \(f(x) = h(g(x))\). This is an essential concept for creating complex functions.

When you compose functions, you must consider the domain and range carefully. The output of the inner function \(g(x)\) must lie within the domain of function \(h(x)\).
  • For example, if \(g(x) = x^2 + 4\) and \(h(x) = \tan^{-1}(x)\), then the composition \(f(x) = \tan^{-1}(x^2 + 4)\) takes the square of \(x\) plus four and finds the inverse tangent of the result.
The careful construction and manipulation of composed functions are integral to calculus and help illuminate the dynamic relationships between different quantities.

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Most popular questions from this chapter

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