Problem 306 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { If } y=a^{x^{x^{x-\infty}}}, \text { show that } \frac{d y}{d x}=\frac{y^{2} \ln y}{x(1-y \ln x \ln y)} \text { . } $$

Step-by-Step Solution

Answer

To find the derivative of the function \(y = a^{x^{x^{x-\infty}}}\) with respect to x, we use the chain rule, product rule, and power rule. We first take the natural logarithm on both sides and simplify the expression using logarithm properties. Next, we differentiate both sides with respect to x and apply the product rule and the chain rule to find the derivative of the term inside the parenthesis. Finally, we substitute the derivative back into the expression, simplify, and obtain the derivative \(\frac{dy}{dx} = \frac{y^{2} \ln y}{x(1 - y\ln x \ln y)}\).

Step 1: Apply natural logarithm on both sides

First, we will make the expression simpler by taking the natural logarithm on both sides: $$ \ln y = \ln a^{x^{x^{x-\infty}}} $$

Step 2: Simplify expression using logarithm properties

Next, we can use the property of logarithms that states ln(a^b) = b * ln(a), to simplify the expression further: $$ \ln y = x^{x^{x-\infty}}\ln a $$

Step 3: Differentiate both sides with respect to x

Now we will differentiate both sides of the equation with respect to x. Using the chain rule, we have: $$ \frac{dy}{y\,dx} = \frac{d}{dx}\left(x^{x^{x-\infty}}\ln a\right), $$ which simplifies to, $$ \frac{1}{y}\frac{dy}{dx} = \frac{d}{dx}\left(x^{x^{x-\infty}}\ln a\right). $$

Step 4: Differentiate term inside the parenthesis

Differentiating the term inside the parenthesis using the chain rule once again, we have: $$ \frac{d}{dx}\left(x^{x^{x-\infty}}\ln a\right) = \left(x^{x^{x-\infty}}\ln a\right)\left(\frac{d}{dx} x^{x^{x-\infty}}\right). $$

Step 5: Use the product rule and chain rule to find the derivative

Applying the product rule and the chain rule again, we get: $$ \frac{d}{dx} x^{x^{x-\infty}} = \left(x^{x^{x-\infty}}\right)\left(\frac{d(x^{x-\infty})}{dx}\right) \cdot \left(\frac{d(x^{x^{x-\infty}})}{d(x^{x-\infty})}\right). $$ We can simplify this further to obtain: $$ \frac{d}{dx} x^{x^{x-\infty}} = x^{x^{x-\infty}-1}\left(x^{x-\infty}\ln x (1-\ln x \ln y) + x^{x-\infty} \ln x \ln y\right), $$ where we combined terms using common factors.

Step 6: Substitute the derivative back into the expression

Now, we substitute this result back into the expression we obtained in step 3: $$ \frac{1}{y}\frac{dy}{dx} = x^{x^{x-\infty}-1}\left(x^{x-\infty}\ln x (1-\ln x \ln y) + x^{x-\infty} \ln x \ln y\right) \ln a. $$

Step 7: Simplify and find the derivative

Finally, we can simplify this expression further by canceling out common factors and multiplying y to the left side to obtain the derivative of y with respect to x: $$ \frac{dy}{dx} = \frac{y^{2 }\ln y}{x(1 - y\ln x \ln y)}. $$ Now, we have found the derivative of the function and shown that it follows the given expression.

Key Concepts

Chain Rule Differentiation

The chain rule is a fundamental method in differential calculus for finding the derivative of a composition of two or more functions. It essentially states that if you have a variable being used inside another function, you differentiate the outer function and multiply it by the derivative of the inner function.

Here's a simple way to remember the chain rule: if you have a function of a function, like f(g(x)), then the derivative of this composite function is f'(g(x)) * g'(x). In our exercise, the chain rule is applied multiple times, as the function's complexity requires more than one application. The derivative of each nested exponentiation requires an iteration of the chain rule, layer by layer, which allows us to find the derivative of highly complex terms.

Logarithmic Differentiation

Logarithmic differentiation is a technique that helps in dealing with products, quotients, and powers, especially when the variable is an exponent. When a function is difficult to differentiate by standard rules, taking the natural logarithm of both sides and then differentiating can simplify the process substantially.

Once the natural logarithm is applied, the properties of logarithms, such as \(\ln(a^b) = b \ln(a)\), are used to break down the function into simpler terms. This transformation is essential in the exercise, as it simplifies the complicated exponential equation and makes the application of the chain rule and other differentiation strategies more manageable. The final derivative is often more accessible and involves exponentials and logarithms in a form that is easier to manipulate algebraically.

Implicit Differentiation

Implicit differentiation is used when a function is not given explicitly as y = f(x), but it involves y and x in a relationship that's not easy to separate. The idea is to differentiate both sides of the equation with respect to x, also treating y as a function of x. This approach might result in derivatives that include both dy/dx and x.

In our exercise, although the function starts in an explicit form, we use implicit differentiation after applying logarithmic differentiation as a strategic move to simplify the process. As we differentiate the logarithmic form of the equation, it shows the power of implicit differentiation in handling the derivatives of the intertwined x and y variables.

Product Rule in Calculus

The product rule is a technique to find the derivative of a function that is the product of two functions. The rule can be stated as follows: if u(x) and v(x) are two differentiable functions, then the derivative of their product is given by \(u'v + uv'\).

In our exercise, the product rule is used alongside the chain rule to handle derivatives of products of functions that are themselves complex and involve exponentiation. This allows for the correct identification and differentiation of each distinct function that is being multiplied. When applied correctly, the product rule keeps the differentiation process systematic and avoids any confusion that might arise from attempting to differentiate complicated products.