Problem 264 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } y=\tan ^{-1} x, \text { find }\left(\frac{d^{2} y}{d x^{2}}\right)_{x=1} $$

Step-by-Step Solution

Answer

The second derivative of the given function at \(x=1\) is \(-\frac{1}{2}\).

Step 1: Find the first derivative of y with respect to x

Since \(y = \tan^{-1}x\), using the chain rule, the first derivative with respect to x is given by: \(\frac{dy}{dx} = \frac{d(\tan^{-1}x)}{dx}\) The derivative of the arctangent function is well-known to be: \(\frac{d(\tan^{-1}x)}{dx} = \frac{1}{1 + x^2}\) So, the first derivative is: \(\frac{dy}{dx} = \frac{1}{1 + x^2}\)

Step 2: Find the second derivative of y with respect to x

Now we need to find the second derivative of y with respect to x, which is the derivative of \(\frac{dy}{dx}\) with respect to x: \(\frac{d^2y}{dx^2} = \frac{d\left(\frac{1}{1 + x^2}\right)}{dx}\) To differentiate this function, we'll use the quotient rule, which gives: \(\frac{d(\frac{u}{v})}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}\) In this case, \(u = 1\) and \(v = 1 + x^2\) Hence, \(\frac{du}{dx} = 0\) (since u is a constant), \(\frac{dv}{dx} = 2x\), Using the quotient rule: \(\frac{d^2y}{dx^2} = \frac{(1 + x^2)\cdot 0 - 1\cdot (2x)}{(1+x^2)^2} = \frac{-2x}{(1+x^2)^2}\)

Step 3: Evaluate the second derivative at x=1

Now, we need to find the value of the second derivative at \(x = 1\). Plug the value \(x=1\) into the expression for the second derivative: \(\left(\frac{d^2y}{dx^2}\right)_{x=1} = \frac{-2(1)}{(1 + (1)^2)^2} = \frac{-2}{(1+1)^2} = \frac{-2}{4} = -\frac{1}{2}\) So, the second derivative of the given function at \(x=1\) is \(-\frac{1}{2}\).

Key Concepts

Arctangent Function

The arctangent function, often written as \( \tan^{-1} x \) or \( \arctan(x) \), is one of the inverse trigonometric functions. It's the inverse of the tangent function. While the tangent function takes an angle and gives its tangent value, the arctangent function takes a tangent value and returns the corresponding angle. Inverse trig functions are special because they help in finding angles from given trigonometric ratios. In the case of \( \arctan(x) \), the input \( x \) can be any real number, and the output is an angle in the range \(-\frac{\pi}{2} \) to \( \frac{\pi}{2} \). Knowing the derivative of the arctangent function is crucial for solving calculus problems. The derivative of \( \tan^{-1} x \) is:

• \( \frac{d}{dx} [\tan^{-1} x ] = \frac{1}{1+x^2} \)

This derivative is used to find other derivatives, such as higher-order derivatives, and solve more complex mathematical problems.

Quotient Rule

The Quotient Rule is a method used in calculus to find the derivative of a function that is the ratio of two differentiable functions. It's especially useful when dealing with a function in the form \( \frac{u}{v} \), where both \( u \) and \( v \) are functions of \( x \). The rule states:

• For \( \frac{u}{v} \), the derivative is \( \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \).

This might seem a bit complex, but it becomes manageable with practice. Consider \( u = 1 \) and \( v = 1 + x^2 \):

• \( \frac{du}{dx} = 0 \) since \( u \) is a constant.
• \( \frac{dv}{dx} = 2x \) because \( v = 1 + x^2 \).

By applying these into the quotient rule formula, you get the derivative of the function, which is essential for computing second derivatives or more.

Second Derivative

In calculus, the second derivative of a function is the derivative of the derivative of that function. While the first derivative gives us information about the rate of change or the slope of the function, the second derivative provides insight into how that rate of change itself is changing. It can indicate concavity and points of inflection on a graph.For the function \( y = \tan^{-1} x \), we first found the first derivative to be \( \frac{1}{1 + x^2} \), as derived from the arctangent function. To find the second derivative, we needed to differentiate \( \frac{1}{1 + x^2} \) again. Using the Quotient Rule:

• We find \( \frac{d^2y}{dx^2} = \frac{-2x}{(1+x^2)^2} \).

Finally, evaluating this at a specific point, like \( x = 1 \), tells us about the curvature at that point. In this case, \( \left( \frac{d^2 y}{dx^2} \right)_{x=1} = -\frac{1}{2} \), meaning the function is concave downwards at \( x = 1 \). Understanding the second derivative is key for analyzing the overall behavior of functions.