Problem 136 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ y=\ln \left(x \sin x \sqrt{1-x^{2}}\right) $$

Step-by-Step Solution

Answer

The derivative of the function \(y=\ln \left(x \sin x \sqrt{1-x^{2}}\right) \) is \(\frac{1}{x \sin x \sqrt{1-x^{2}}} \cdot (\sin x \sqrt{1-x^{2}} + x \cos x \sqrt{1-x^{2}} - x^2 \sin x /( \sqrt{1- x^{2}}))\).

Step 1: Identify the outer function

In this case, the outermost function is the natural logarithm, \(\ln\). The derivative of \(\ln(u)\) where \(u\) is any given function, is \(\frac{1}{u} \cdot u'\). Here, \(u=x \sin x \sqrt{1-x^{2}}\).

Step 2: Differentiate the natural logarithm function

Using the derivative rule for the natural logarithm, the derivative of \(y\) is \( \frac{1}{x \sin x \sqrt{1-x^{2}}} \cdot (x \sin x \sqrt{1-x^{2}})'\)

Step 3: Identify the inner function

The inner function here can be considered as \(v=x \sin x \sqrt{1-x^{2}}\). This function is a product of three functions (\(x\), \(\sin x\), and \(\sqrt{1-x^{2}}\)). So it will be differentiated using the product rule. The product rule states that the derivative of two multiplied functions is the first function times the derivative of the second function, plus the second function times the derivative of the first function.

Step 4: Differentiate the inner function

The derivative of \(v\) is \(x' \sin x \sqrt{1-x^{2}} + x \sin x' \sqrt{1-x^{2}} + x \sin x \sqrt{1-x^{2}}'\). Simplifying, \(v' = \sin x \sqrt{1-x^{2}} + x \cos x \sqrt{1-x^{2}} - x^2 \sin x /( \sqrt{1- x^{2}})\)

Step 5: Substitute back

Finally, we substitute the derivative of \(v\) into our earlier equation from Step 2. The final derivative of the function is \(\frac{1}{x \sin x \sqrt{1-x^{2}}} \cdot (\sin x \sqrt{1-x^{2}} + x \cos x \sqrt{1-x^{2}} - x^2 \sin x /( \sqrt{1- x^{2}}))\)

Key Concepts

Product Rule

When it comes to calculus, the product rule is an essential tool for finding the derivative of a product of two functions. It is formally stated as \( (f(x) \cdot g(x))' = f'(x) \cdot g(x) + f(x) \cdot g'(x) \). Simply put, it means that to differentiate a product, you first take the derivative of the first function and multiply it by the second unchanged function, and then add the first function multiplied by the derivative of the second.

In the provided exercise, we apply the product rule to the inner function \(v = x \sin x \sqrt{1-x^{2}}\). Here, \(v\) is the product of three functions which complicates things slightly. In such cases, you can extend the product rule: differentiate the first function, keep the other two as they are, and then continue this process circle-wise for the other functions involved. This step is where the exercise incorporates the product rule to simplify the seemingly complex inner function.

Chain Rule

The chain rule is another indispensable concept in calculus, particularly when dealing with composite functions. It allows you to find the derivative of a function that is composed of other functions. The formula for the chain rule is \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) where \(y\) is a function of \(u\) and \(u\) is a function of \(x\). In other words, to differentiate a composite function, you first differentiate the outer function with respect to the inner function, and then multiply this by the derivative of the inner function with respect to \(x\).

Our exercise has \(y = \ln( v)\), making \(v\) the inner function. Using the chain rule, we differentiate the outer natural logarithm function with respect to \(v\), and then multiply by the derivative of \(v\) with respect to \(x\), which has already been calculated using the product rule.

Derivatives of Trigonometric Functions

Trigonometric functions such as sine, cosine, and tangent have specific derivative rules. The derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \( -\sin x \). These derivatives often appear in calculus problems involving periodic functions or oscillating motion.

The incorporation of trigonometric functions into the exercise at hand is precisely what triggers the intricacy of the product rule application. We have the function \(x \sin x \), which after differentiation gives \(x \cos x \) as part of the derivative for the expression within the natural logarithm. Understanding the derivative patterns for basic trigonometric functions is crucial for correctly simplifying and differentiating more complex expressions like the one in our exercise.

Implicit Differentiation

Implicit differentiation is a technique used when a function is not given explicitly as \(y=f(x)\) but rather in a form where \(y\) and \(x\) are intermingled. Even though our exercise does not require implicit differentiation, it's important to know that this method involves differentiating both sides of an equation with respect to \(x\) while applying the chain rule when dealing with \(y\), as \(y\) is also a function of \(x\). The process often results in having \(dy/dx\) terms in your equation, which you then solve for to find the derivative.

For example, if you had the equation \( xy + \sin y = 1\), implicitly differentiating with respect to \(x\) while treating \(y\) as an implicit function of \(x\) would give you \( y + x(dy/dx) + \cos y (dy/dx) = 0\). Even though we're not using implicit differentiation here, mastering it would allow you to tackle broader types of differentiation problems.