Problem 157 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ y=(\tan 2 x)^{\cot ^{\frac{x}{2}}} $$

Step-by-Step Solution

Answer

The derivative of the function \(y = (\tan 2x)^{\cot^{\frac{x}{2}}}\) is: \(\frac{dy}{dx} = (\tan 2x)^{\cot^{\frac{x}{2}}} * (\frac{2\sec^2{2x}}{\tan{2x}}\cot^{\frac{x}{2}} - (\ln(\tan 2x))\frac{\cos x}{\sin^2{\frac{x}{2}}})\)

Step 1: Rewrite in Exponential Form

Rewrite the function in exponential form using the property: \(a^b = e^{(\ln a) * b}\) \(y = e^{\ln(\tan 2x) * \cot^{\frac{x}{2}}}\) Now, we can find the derivative of the function.

Step 2: Differentiate with Respect to x

We will use the chain rule to differentiate: \(\frac{dy}{dx} = e^{\ln(\tan 2x) * \cot^{\frac{x}{2}}} * \frac{d}{dx} [\ln(\tan 2x) * \cot^{\frac{x}{2}}]\) Now, we will find the derivative of the expression \(\ln(\tan 2x) * \cot^{\frac{x}{2}}\).

Step 3: Use Product Rule for Differentiation

To find the derivative of the expression, we'll use the product rule: \(u = \ln(\tan 2x)\) and \(v = \cot^{\frac{x}{2}}\) \(\frac{d}{dx}[uv] = u'v + uv'\) First, we'll find \(u'\) and \(v'\): \(u' = \frac{d}{dx} [\ln(\tan 2x)] = \frac{1}{\tan 2x} * \frac{d}{dx} [\tan 2x]\) \(u' = \frac{1}{\tan 2x} * (2 * \sec^2{2x})\) \(u' = \frac{2\sec^2{2x}}{\tan{2x}}\) Similarly, we'll find \(v'\): \(v' = \frac{d}{dx} [\cot^{\frac{x}{2}}]\) For this, we can rewrite \(\cot^{\frac{x}{2}}\) as \(\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\) \(v' = \frac{d}{dx} [\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}]\) Now, we can apply the quotient rule: \(v' = \frac{(\frac{-1}{2}\sin{\frac{x}{2}})(\sin{\frac{x}{2}}) - (\cos{\frac{x}{2}})(\frac{1}{2}\cos{\frac{x}{2}})}{(\sin^2{\frac{x}{2}})^2}\) \(v' = -\frac{\cos x}{\sin^2{\frac{x}{2}}}\) Now, we can plug \(u'\) and \(v'\) into the product rule formula: \(\frac{d}{dx}[uv] = u'v + uv'\) \(\frac{d}{dx} [\ln(\tan 2x) * \cot^{\frac{x}{2}}] = (\frac{2\sec^2{2x}}{\tan{2x}})(\cot^{\frac{x}{2}}) - (\ln(\tan 2x))\frac{\cos x}{\sin^2{\frac{x}{2}}}\)

Step 4: Combine Results

Finally, combine the results from Steps 2 and 3: \(\frac{dy}{dx} = e^{\ln(\tan 2x) * \cot^{\frac{x}{2}}} * (\frac{2\sec^2{2x}}{\tan{2x}}\cot^{\frac{x}{2}} - (\ln(\tan 2x))\frac{\cos x}{\sin^2{\frac{x}{2}}})\) \(\frac{dy}{dx} = (\tan 2x)^{\cot^{\frac{x}{2}}} * (\frac{2\sec^2{2x}}{\tan{2x}}\cot^{\frac{x}{2}} - (\ln(\tan 2x))\frac{\cos x}{\sin^2{\frac{x}{2}}})\) This is the derivative of the given function.

Key Concepts

Chain Rule

When it comes to differentiating complex functions, the chain rule is your go-to tool. Imagine you have a function inside another function, each affected by a variable. To differentiate this kind of nested function, the chain rule helps us unwind the composition. Suppose you have a function like \( f(g(x)) \). Here, \( g(x) \) nests within \( f \). The chain rule tells us how to take the derivative:

• First, take the derivative of the outer function \( f \), keeping the inner function \( g(x) \) unchanged.
• Next, multiply this result by the derivative of the inside function \( g(x) \).

In our example exercise, we used the chain rule to handle the transformation of the function \( y = e^{\ln(\tan 2x) * \cot^{\frac{x}{2}}} \). The outer function \( e^u \) required differentiation, along with the inner, more complex product of \( \ln(\tan 2x) \) and \( \cot^{\frac{x}{2}} \). Distinguishing each part simplifies the process.

Product Rule

The product rule comes into play when you need to differentiate a multiplication of two functions. Imagine two functions, \( u \) and \( v \), that are multiplied together, such as in \( u \, v \). To differentiate \( u \, v \), the product rule guides us with a straightforward method:

• Take the derivative of the first function, \( u' \), and multiply it by the unchanged second function, \( v \).
• Add this to the derivative of the second function, \( v' \), multiplied by the unchanged first function, \( u \).

The formula is neatly summarized as: \( u'v + uv' \). During the example solution, this rule was necessary for differentiating \( \ln(\tan 2x) \cdot \cot^{\frac{x}{2}} \), because it involves the product of the two individual functions. Recognizing when to apply the product rule saves time and prevents errors.

Exponential Functions

Exponential functions, characterized by a variable as the exponent, appear frequently in calculus. A basic exponential form might look like \( a^x \), where \( a \) is a constant base and \( x \) is the variable exponent.Differentiating exponential functions typically results in a function that resembles the original form, often multiplied by a factor that comes from the chain or product rules. For example, differentiating \( e^{g(x)} \) involves:

• First, keeping the exponential \( e^{g(x)} \) unchanged.
• Next, multiply this by \( g'(x) \), the derivative of the exponent function.

In the exercise, transforming the given expression into an exponential form \( y = e^{\ln(\tan 2x) * \cot^{\frac{x}{2}}} \) helped streamline the differentiation process. This step allows for maintaining consistency in the differentiation, especially when exponential functions combine with trigonometric ones. Mastering exponential differentiation equips you to tackle a wide array of mathematical problems effectively.