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What is the slope of the line tangent to the graph of \(y=\tan ^{-1} x\) at \(x=-2 ?\)

Short Answer

Expert verified
Answer: The slope of the tangent line at \(x=-2\) is \(\frac{1}{5}\).

Step by step solution

01

Find the derivative of \(y=\tan ^{-1}x\) with respect to x.

To find this derivative, we will use the chain rule. Since the derivative of \(\tan ^{-1}x\) with respect to \(x\) is \(\frac{1}{1+x^2}\), so the derivative with respect to \(x\) is given by: $$ \frac{dy}{dx}=\frac{1}{1+x^2} $$
02

Evaluate the derivative at \(x=-2\)

Now that we have the derivative of \(y=\tan ^{-1}x\) with respect to \(x\), we will now substitute \(x=-2\) into the equation to find the slope of the tangent at this point: $$ \frac{dy}{dx} \Bigg|_{x=-2} = \frac{1}{1+(-2)^2} $$
03

Simplify

Finally, we will simplify the equation to find the slope of the tangent line at \(x=-2\): $$ \frac{dy}{dx} \Bigg|_{x=-2} = \frac{1}{1+4} = \frac{1}{5} $$ So, the slope of the tangent line to the graph of \(y=\tan^{-1}x\) at \(x=-2\) is \(\frac{1}{5}\).

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Key Concepts

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

Inverse Trigonometric Functions
Inverse trigonometric functions are the inverse operations of the basic trigonometric functions such as sine, cosine, and tangent. They are used to determine the angle when the ratio of the sides is known. In the context of calculus, inverse trigonometric functions help in finding derivatives and analyzing the behavior of functions. The notation \( \tan^{-1} x \) represents the arctangent of \( x \), which gives the angle whose tangent is \( x \).

When calculating the derivative of an inverse trigonometric function, each has a specific formula. For example, the derivative of \( \tan^{-1} x \) with respect to \( x \) is \( \frac{1}{1+x^2} \).

Understanding these derivatives is crucial when finding slopes of tangent lines or integrating functions, as they often appear in various engineering and physics problems.
Chain Rule
The chain rule is a fundamental principle in calculus used to find the derivative of the composition of two or more functions. It states that if a function \( y \) can be expressed as a composition of other functions, say \( y = f(g(x)) \), then the derivative \( \frac{dy}{dx} \) is found as \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).

The chain rule is particularly useful when dealing with inverse trigonometric functions, which typically involve more complex expressions within their derivatives or when they are part of larger composite functions. This rule helps simplify the process, making it easier to differentiate.

While working with derivatives like \( \frac{dy}{dx} = \frac{1}{1+x^2} \), even if it seems straightforward, understanding the underlying principle, like the chain rule, ensures accuracy and a deeper comprehension of mathematical functions.
Tangent Line Slope
The slope of the tangent line to a curve at a given point gives us the rate of change of the function at that point. It is essentially the derivative of the function evaluated at that specific value of \( x \).

To calculate this, we first differentiate the function with respect to \( x \), and then substitute the given \( x \)-value into this derivative. For example, the function \( y = \tan^{-1} x \) has a derivative of \( \frac{dy}{dx} = \frac{1}{1+x^2} \).

This expression is then evaluated at the desired point, as shown in the given solution at \( x = -2 \), leading to a slope of \( \frac{1}{5} \). The tangent line's slope tells us how steep the line is and the direction in which it inclines or declines at the given point on the graph.

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Most popular questions from this chapter

Jean and Juan run a one-lap race on a circular track. Their angular positions on the track during the race are given by the functions \(\theta(t)\) and \(\varphi(t),\) respectively, where \(0 \leq t \leq 4\) and \(t\) is measured in minutes (see figure). These angles are measured in radians, where \(\theta=\varphi=0\) represent the starting position and \(\theta=\varphi=2 \pi\) represent the finish position. The angular velocities of the runners are \(\theta^{\prime}(t)\) and \(\varphi^{\prime}(t)\). a. Compare in words the angular velocity of the two runners and the progress of the race. b. Which runner has the greater average angular velocity? c. Who wins the race? d. Jean's position is given by \(\theta(t)=\pi t^{2} / 8 .\) What is her angular velocity at \(t=2\) and at what time is her angular velocity the greatest? e. Juan's position is given by \(\varphi(t)=\pi t(8-t) / 8 .\) What is his angular velocity at \(t=2\) and at what time is his angular velocity the greatest?

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