Problem 130 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ y=e^{\tan ^{-1} \sqrt{1+\ln (2 x+3)}} $$

Step-by-Step Solution

Answer

The derivative of \(y=e^{\tan ^{-1} \sqrt{1+\ln (2 x+3)}}\) with respect to \(x\) will be: \(\frac{e^{\tan ^{-1} \sqrt{1+\ln (2 x+3)}}}{(1+(\sqrt{1 + \ln(2x+3)})^2)(\sqrt{1 + \ln(2x+3)})(2x+3)} \times \frac{1}{2 \sqrt{1+\ln(2x+3)}} \times 2\).

Step 1: Differentiation of the Outermost Function

The outermost function is \(y=e^u\) where \(u=\tan ^{-1} \sqrt{1+\ln (2 x+3)}\). The derivative of \(e^u\) with respect to \(x\) is \(e^u \times \frac{du}{dx}\). So let's find \(du/dx\).

Step 2: Differentiation of the inner function

Here, \(u=\tan ^{-1} v\) where \(v= \sqrt{1+\ln (2 x+3)}\). The derivative of \(\tan ^{-1} v\) with respect to \(x\) is \(\frac{1}{1+v^2}\times \frac{dv}{dx}\). Now let's find \(dv/dx\).

Step 3: Differentiate \(v = \sqrt{w}\)

In this stage, \(v=\sqrt{w}\) where \(w=1+\ln (2 x+3)\). The derivative of \(\sqrt{w}\) with respect to \(x\) is \(\frac{1}{2 \sqrt{w}} \times \frac{dw}{dx}\). Now let's calculate \(dw/dx\).

Step 4: Differentiate \(w = 1 + \ln(z)\)

Here, \(w=1+\ln(z)\) and \(z=2x+3\). The derivative of \(1+\ln(z)\) with respect to \(x\) is \(\frac{1}{z} \times \frac{dz}{dx}\). The derivative of \(2x+3\) with respect to \(x\) is 2.

Step 5: Combine Updates and Simplify

Combine all the derivatives accordingly from step 1 to 4, and simplify.

Key Concepts

Chain Rule

The chain rule is a fundamental concept in differential calculus used to differentiate composite functions. It applies when you have a function composed of multiple layers, each nested inside another.
For example, consider a function, say \( y = f(g(x)) \), here \( g(x) \) is nested within \( f \). The chain rule helps in differentiating such functions.
The essence of the chain rule is that you take the derivative of the outer function and multiply it by the derivative of the inner function.
Mathematically, it is expressed as:

• 
  \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

In our exercise, this method is used to differentiate the complex function \( y=e^{\tan ^{-1} \sqrt{1+\ln(2x+3)}} \). This involves differentiating each layer step by step. Thus, understanding the chain rule ensures a systematic approach, breaking down complex differentiation into manageable parts.

Exponential Functions

Exponential functions are characterized by the constant \( e \), which is approximately 2.71828, often referred to as Euler's number. They occur frequently in calculus due to their unique properties.
The exponential function \( e^x \), where the base is \( e \) and the exponent is \( x \), has the distinct property that its derivative \( \frac{d}{dx}e^x = e^x \). This self-replicating feature makes exponential functions simple yet powerful in differential calculus.
In our exercise, the outermost function is an exponential function: \( y = e^u \), and it illustrates the application of the chain rule in conjunction with exponential differentiation.

• The derivative, when combined with the chain rule gives:
  \( \frac{dy}{dx} = e^u \cdot \frac{du}{dx} \)

This showcases the elegance and necessity of understanding exponential functions and their derivatives when tackling complex calculus problems.

Inverse Trigonometric Functions

Inverse trigonometric functions reverse the process of their respective trigonometric functions. They are essential in calculus for dealing with angles and converting between functions.
Common inverse trigonometric functions include \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), otherwise known as arcsin, arccos, and arctan.
The derivative of the inverse tangent function, specifically \( \tan^{-1}(x) \), is crucial in many calculus problems. It follows the derivative formula:

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

In our exercise, \( \tan^{-1}(\sqrt{1 + \ln(2x+3)}) \) is an intermediate step. Understanding how to deal with such inverse functions is key to unlocking the differentiation of composite functions.

Logarithmic Differentiation

Logarithmic differentiation is a technique that simplifies differentiating complex functions by using logarithms. This approach is particularly useful when dealing with products, quotients, or powers.
By taking the natural logarithm of a function, you convert complex multiplication or division into more manageable addition or subtraction problems via properties of logarithms.
For instance, the natural log of a product \( \ln(uv) \), becomes

• \( \ln(u) + \ln(v) \)

, simplifying the differentiation process.
In our specific example, logarithmic techniques come into play with the component \( \ln(2x+3) \). Differentiating this with respect to \( x \) results in:

• \( \frac{d}{dx} \ln(2x+3) = \frac{1}{2x+3} \cdot 2 \)

Such clarity in differentiation is made possible by leveraging the power of logarithms.