Question

Multiple Choice Which of the following is the slope of the tangent line to \(y=\tan ^{-1}(2 x)\) at \(x=1 ?\) \(\begin{array}{llll}{\text { (A) }-2 / 5} & {\text { (B) } 1 / 5} & {\text { (C) } 2 / 5} & {\text { (D) } 5 / 2}\end{array}\) \((\mathbf{E}) 5\)

Short answer

The correct choice is (C) \( \frac{2}{5} \). This is the slope of the tangent line to the curve \(y=\tan ^{-1}(2 x)\) at \(x=1\).

Step-by-step solution

Differentiation

Differentiate the function \(y = \tan^{-1}(2x)\) using the chain rule. The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1 + x^2} \), and the derivative of \(2x\) is \(2\). So, the derivative becomes \( y' = \frac{2}{1 + (2x)^2} \)

Substitute x = 1

Substitute \( x = 1 \) into the derivative, we get \( y' = \frac{2}{1 + (2*1)^2} = \frac{2}{5} \)

Identify the right option

Looking at the given options, it is evident that option (C) \( \frac{2}{5} \) is the correct choice.

Key concepts

Differentiation and Tangent Line Slope

Differentiation is a fundamental concept in calculus, primarily concerned with finding the rate at which one quantity changes with respect to another. When it comes to the slope of a tangent line to a curve at a given point, differentiation takes the spotlight.
The slope of the tangent line represents the instantaneous rate of change of the function at that point. It's like capturing a snapshot of an object's speed at a specific moment in time. In the case of the curve defined by the function y = tan^{-1}(2x), finding the slope involves taking the derivative of this function.
To envision this, think of drawing a line that just skims the curve at the point of tangency, x=1, in our case. The gradient or steepness of this line is exactly what we determine through differentiation. By calculating the derivative of the function and evaluating it at the given point, we obtain the precise inclination of the tangent line; thus allowing us to summarize the curve's local behavior concisely.

Chain Rule

Like a brilliant shortcut in a maze, the chain rule in calculus is an efficient way to differentiate composite functions – functions made up of two or more functions, nested within each other. This rule can be intriguing; imagine it like peeling an onion, layer by layer, to comprehend the change in each part.
The formula for the chain rule is expressed as: (f(g(x)))' = f'(g(x))g'(x). Here, f represents the outer function and g the inner function.

Applying the Chain Rule

In our exercise's context, to find the derivative of y = tan^{-1}(2x), we consider tan^{-1}(x) as the outer function, f, and 2x as the inner function, g. Following the onion analogy, we first differentiate the outer layer, tan^{-1}(x), yielding 1/(1+x^2), and then we tackle the inner layer, which is straightforward since the derivative of 2x is just 2.
We combine these partial derivatives according to the chain rule, resulting in the composite derivative y' = 2/(1+(2x)^2). The chain rule shows its convenience as it allows us to untangle the rate of change in complex, related quantities.

Inverse Trigonometric Function Derivatives

Diving deeper into differentiation, we encounter the derivatives of inverse trigonometric functions, which can sometimes be as enigmatic as deciphering ancient scripts. These functions allow us to unwind angles from given trigonometric values.
Take for instance the function y = tan^{-1}(x); it gives us the angle whose tangent is x. When differentiating inverse trigonometric functions, the outcomes are not merely straightforward constants or polynomial terms; instead, they provide ratios that unveil the original function's rate of change.

Key Derivatives:

• The derivative of tan^{-1}(x) is 1/(1+x^2).
• The derivative of sin^{-1}(x) or cos^{-1}(x) incorporates the square root of 1 - x^2.

By acknowledging these unique derivatives, we can deftly navigate through problems involving curves like the one from our exercise. Here, recognizing that the derivative of tan^{-1}(x) is 1/(1 + x^2) was crucial to finding the slope. Ultimately, a proper grasp of these derivatives is instrumental for smoothly sailing through the challenges of calculus.