Problem 3
Question
What is the slope of the line tangent to the graph of \(y=\tan ^{-1} x\) at \(x=-2 ?\)
Step-by-Step Solution
Verified Answer
Answer: The slope of the tangent line at \(x=-2\) is \(\frac{1}{5}\).
1Step 1: 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:
$$
\frac{dy}{dx}=\frac{1}{1+x^2}
$$
2Step 2: 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:
$$
\frac{dy}{dx} \Bigg|_{x=-2} = \frac{1}{1+(-2)^2}
$$
3Step 3: 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 \(\frac{1}{5}\).
Key Concepts
Inverse Trigonometric FunctionsChain RuleTangent Line Slope
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.
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.
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.
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.
Other exercises in this chapter
Problem 3
Complete the following statement: If \(\frac{d y}{d x}\) is small, then small changes in \(x\) result in relatively _________ changes in the value of \(y\).
View solution Problem 3
Show that \(\frac{d}{d x}(\ln k x)=\frac{d}{d x}(\ln x),\) where \(x>0\) and \(k\) is a positive real number.
View solution Problem 3
Why are both the \(x\) -coordinate and the \(y\) -coordinate generally needed to find the slope of the tangent line at a point for an implicitly defined functio
View solution Problem 3
Fill in the blanks. The derivative of \(f(g(x))\) equals \(f^{\prime}\) evaluated at ________ multiplied by \(g^{\prime}\) evaluated at ________
View solution