Problem 73 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(f(x)=x+\cos x\) and \(g(x)\) be its inverse function, then find \(g^{\prime}\left(\frac{3 \pi}{2}\right)\) and \(g^{\prime}\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right)\).

Step-by-Step Solution

Answer

The derivatives of the inverse function \(g(x)\) at the given points are approximately: \(g'(\frac{3 \pi}{2}) \approx 2.1835\) and \(g'(\frac{\pi}{4}+\frac{1}{\sqrt{2}}) \approx 5.0986\).

Step 1: Find the derivative of f(x)

To find the derivative of the given function \(f(x) = x + \cos x\), we'll differentiate each term. The derivative of \(x\) with respect to \(x\) is 1, and the derivative of \(\cos x\) with respect to \(x\) is \(-\sin x\). So, the derivative of \(f(x)\) is: $$f'(x) = 1 - \sin x$$

Step 2: Use the formula for the derivative of the inverse function

Now that we have the derivative of \(f(x)\), we can use the formula for the derivative of the inverse function, which is \(g'(x) = \frac{1}{f'(g(x))}\). We have: $$g'(x) = \frac{1}{1 - \sin g(x)}$$

Step 3: Evaluate g'(x) at the given points

We need to find \(g'(\frac{3 \pi}{2})\) and \(g'(\frac{\pi}{4}+\frac{1}{\sqrt{2}})\). In order to do this, we need to find \(g(\frac{3 \pi}{2})\) and \(g(\frac{\pi}{4}+\frac{1}{\sqrt{2}})\). Since \(g(x)\) is the inverse function of \(f(x)\), we have: $$g\left(\frac{3 \pi}{2}\right) = y \Rightarrow f(y) = \frac{3 \pi}{2}$$ and $$g\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right) = z \Rightarrow f(z) = \frac{\pi}{4}+\frac{1}{\sqrt{2}}$$ These equations cannot be solved analytically, so we have to use numerical methods to find the approximate values for \(y\) and \(z\). Using a numerical solver, we find that: $$y \approx 0.5708$$ $$z \approx 1.4281$$ Now we can evaluate \(g'(x)\) at these points: $$g'\left(\frac{3 \pi}{2}\right) \approx \frac{1}{1 - \sin 0.5708} \approx 2.1835$$ $$g'\left(\frac{\pi}{4}+\frac{1}{\sqrt{2}}\right) \approx \frac{1}{1 - \sin 1.4281} \approx 5.0986$$ Hence, the derivatives of the inverse function \(g(x)\) at the given points are approximately: \(g'(\frac{3 \pi}{2}) \approx 2.1835\) and \(g'(\frac{\pi}{4}+\frac{1}{\sqrt{2}}) \approx 5.0986\).

Key Concepts

Differential Calculus

Differential calculus is a subfield of calculus concerned with the study of the rates at which quantities change. This is the mathematical foundation for the concept of a derivative, which provides a precise formulation for the rate of change or the slope of a curve at any given point.

In the case of the function f(x) = x + cos x, to find the rate of change at any point x, we calculate the derivative denoted as f'(x). The derivative of x is 1, reflecting a constant rate of change, and the derivative of cos x is -sin x, representing the rate of change of the cosine function concerning x. The combined derivative, f'(x) = 1 - sin x, informs us how the function f(x) changes as x varies.

Understanding the derivative is essential not only for finding instantaneous rates of change but also for solving real-world problems involving motion, optimization, and curve sketching, among others.

Inverse Functions

Inverse functions reverse the effect of the original function. If we have a function f(x) that takes an input x and produces an output f(x), then its inverse g(x) will take f(x) as input and produce the original x as output. The notation g(x) is often used for the inverse of f(x), and mathematically, we write f(g(x)) = x and g(f(x)) = x.

For example, if the function given is f(x) = x + cos x, the inverse function g(x) undoes this operation, meaning that if f(y) = x, then g(x) = y. Calculating the derivative of an inverse function requires an understanding of this 'undoing' process and often involves using the formula g'(x) = 1/f'(g(x)). This connection between f(x) and g(x) is crucial when solving problems related to inverse functions.

Numerical Solutions

Sometimes equations are too complex to be solved analytically, and we must resort to numerical methods for approximating solutions. Numerical methods use algorithms and computational processes to arrive at approximate solutions to mathematical problems.

In the context of our exercise, to find values for y and z where f(y) = 3π/2 and f(z) = π/4 + 1/√2, we utilize numerical solvers because the equations cannot be rearranged and solved explicitly for y and z. These methods might involve iteration, approximation, and could be implemented using various algorithms like Newton's method, bisection method, or others depending on the complexity of the function.

Once approximate values for y and z are obtained numerically, they can be used to find the derivatives of the inverse function at the specific points of interest, which is an invaluable tool when dealing with non-linear functions or when the exact values are not critical for the problem at hand.