Question

If \(y=(x \log x)^{\log \log x}\) for \((d y / d x)=A^{\log \log x-1}(B+(\log x+2) \log\) \(\log x\) ) then \(A B=\) (a) \(x \log x\) (b) \(x(\log x)^{2}\) (c) \(x^{2} \cdot \log x\) (d) \(\log x\)

Short answer

The value of AB is \(x(\log x)^{2}\), which corresponds to option (b).

Step-by-step solution

Differentiate y with respect to x

To find the derivative \(\frac{dy}{dx}\), we will apply both the chain rule and the product rule. The chain rule states that if we have a composite function, then its derivative is the product of the derivative of the outer function applied to the inner function and the derivative of the inner function. Here, our outer function is \(u^{\log \log x}\) and our inner function is \(u = x \log x\). Using the chain rule, \[\frac{dy}{dx}=\frac{d}{dx}\Big((x \log x)^{\log \log x}\Big) = (\log \log x)(x \log x)^{\log \log x -1} \frac{d}{dx}(x \log x)\]

Differentiate the inner function with respect to x

The inner function is \(u = x \log x\). To differentiate this, we will use the product rule, which states that the derivative of the product of two functions is the sum of the product of the first function and the derivative of the second function and the product of the second function and the derivative of the first function. Using the product rule, we have: \[\frac{d}{dx}(x \log x) = x \frac{d}{dx}(\log x) + \log x \frac{d}{dx}(x)\] Now differentiate each term: \[\frac{d}{dx}(x \log x) = x \frac{1}{x} + \log x (1) = 1 + \log x\]

Plug ∂(x ∙ log(x))/∂x back into dy/dx

Now substitute back the value of \(\frac{d}{dx}(x \log x)\) in the derivative of the main function: \[\frac{dy}{dx} = (\log \log x)(x \log x)^{\log \log x -1} (1 + \log x)\]

Compare with given derivative to find A and B

Given derivative is: \(\frac{dy}{dx} = A^{\log \log x - 1}(B+(\log x + 2)\log \log x)\). By comparing it with the derivative we found above, we can equate the two expressions: \[(\log \log x)(x \log x)^{\log \log x -1} (1 + \log x) = A^{\log \log x - 1}(B+(\log x + 2)\log \log x)\] Now comparing the two sides, we can conclude that: \(A = (x \log x)\) and \(B = (\log x + 2) \log \log x - \log x\)

Find the value of AB

To find the value of AB, we just multiply A and B. \[AB = (x \log x)\Big((\log x + 2) \log \log x - \log x\Big)\] Looking for a matching option, we find that this expression corresponds to option (b), which is \(x(\log x)^{2}\). Therefore, the answer is (b).

Key concepts

Chain Rule

The chain rule is a fundamental method used in calculus when dealing with composite functions. In simpler terms, a composite function is a function made up of other functions. The chain rule helps us find the derivative of such functions.

Consider a composite function like \( f(g(x)) \), where \( g(x) \) is the inner function and \( f \) is the outer function. The chain rule states that the derivative of this composite function is:

• The derivative of the outer function, evaluated at the inner function, multiplied by the derivative of the inner function.

In mathematical terms, this is given by:\[(f \, ext{∘} \, g)'(x) = f'(g(x)) \cdot g'(x)\]Using the chain rule helps simplify complex differentiation tasks, letting you take derivatives in steps rather than all at once, which is especially useful for expressions like \((x \log x)^{\log \log x}\).

In our example, the chain rule was used to differentiate \((x \log x)^{\log \log x}\) by treating \(x \log x\) as the inner function and \(u^{\log \log x}\) as the outer function. This clear separation into parts allows you to handle each derivative calculation in stages, a highly manageable process for complex expressions.

Product Rule

The product rule is another essential tool in calculus, designed to deal with the differentiation of products of two functions. When applying the product rule, you take into account both functions and their individual derivatives.

Given two functions, say \( u(x) \) and \( v(x) \), the product rule states that the derivative of their product is:

• The derivative of the first function times the second, plus the first function times the derivative of the second.

Mathematically, this can be written as:\[(uv)' = u'v + uv'\]In our exercise, we see this rule applied directly to differentiate \(x \log x\), where:

• \(u(x) = x\) and \(v(x) = \log x\),

leading to:

• \( u' = 1 \) and \( v' = \frac{1}{x} \)
• The derivative: \( x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x \).

Understanding when and how to apply the product rule is crucial for correctly handling derivatives of multiplied functions, making it a practical skill for breaking down multifaceted mathematical tasks.

Logarithmic Functions

Logarithms turn multiplication into addition and are the inverse of exponential functions. They simplify complex expressions and are pivotal in calculus, especially when differentiating functions that involve exponents or products.

A logarithmic function is presented as \( \log_b(x) \), where \( b \) is the base, and it denotes the power to which the base must be raised to produce the number \( x \). In many mathematical scenarios, the natural logarithm \( \ln(x) \) or \( \log_e(x) \) is commonly used, simplifying derivatives due to its friendly properties.

For instance, some useful logarithmic rules include:

• \( \log(ab) = \log(a) + \log(b) \)
• \( \log\left(\frac{a}{b}\right) = \log(a) - \log(b) \)
• \( \log(a^c) = c \cdot \log(a) \)

These properties are incredibly helpful when differentiating expressions like \( x \log x \). In calculus, especially when dealing with complex power functions like \( (x \log x)^{\log \log x} \), transforming them with logarithms can simplify differentiation tasks.

Here, separating components such as \( (x \log x)^{\log \log x} \) into more manageable parts using these rules allows for easier application of differentiation techniques.

Derivatives of Composite Functions

When faced with composite functions, which are functions within functions, differentiation requires a strategic approach. This involves using the chain rule to simplify the complexity.

In calculus, composite functions often appear in expressions like \( f(g(x)) \), meaning one function is nested inside another. Deriving such functions is about carefully unwinding them step by step.

Here's what you typically do when finding the derivative of these functions:

• Identify the outer and inner functions.
• Use the chain rule: differentiate the outer, evaluate at the inner, then multiply by the derivative of the inner.

For example, in our exercise with \((x \log x)^{\log \log x}\), focus on the whole expression as a blend of functions:

• The power function as the outer function.
• \(x \log x\) as the inner function.

Understanding each function's role streamlines the differentiation process. The outer examines the power dependency on the inner, while the inner gives the base's input value. This method, valuably streamlined using the chain rule, ensures that even the most entwined functions can surrender their complexities, making derivatives approachable and solvable.