Question

If y=logⁿx, where logⁿ means log_e log_e log_e .....(repeated n times), then xlogx log²x log³x....log ^n-1 xlogⁿx \(\frac{dy}{dx}\) is equal to :

• A. log x
• B. x
• C. 1
• D. logⁿx

Answer 1

The Correct Option is D

Given: \( y = \log^n x = (\log x)^n \)

Differentiating with respect to \( x \):
\[ \frac{dy}{dx} = n (\log x)^{n-1} \cdot \frac{1}{x} \] 

Now multiply both sides by:
\[ x \cdot \log x \cdot (\log x)^2 \cdot (\log x)^3 \cdots (\log x)^{n-1} \cdot (\log x)^n = x \cdot (\log x)^{1+2+\cdots+n} \]

After simplification, remaining expression is:
\[ \log^n x \]

Final Answer: Option (D): \( \log^n x \)

Answer 2

Let's define the repeated logarithm function as follows: 

\(log_1 x = log(x)\)

\(log_2 x = log(log(x))\)

\(log_3 x = log(log(log(x)))\)

And so on, until

\(log_n x = log(log_{n-1}(x))\)

We are given \(y = log_n x\), which means \(y = log(log_{n-1}(x))\).

We need to find \(\frac{dy}{dx}\). Using the chain rule, we have:

\(\frac{dy}{dx} = \frac{d}{dx} (log(log_{n-1}(x))) = \frac{1}{log_{n-1}(x)} \cdot \frac{d}{dx} (log_{n-1}(x))\)

Let's apply this repeatedly. For example:

\(\frac{d}{dx}log_1 x = \frac{1}{x}\)

\(\frac{d}{dx}log_2 x = \frac{d}{dx}log(log(x)) = \frac{1}{log(x)} \cdot \frac{1}{x}\)

\(\frac{d}{dx}log_3 x = \frac{d}{dx}log(log(log(x))) = \frac{1}{log(log(x))} \cdot \frac{1}{log(x)} \cdot \frac{1}{x}\)

In general, for \(y=log_n x\), we have:

\(\frac{dy}{dx} = \frac{1}{x \cdot log(x) \cdot log_2(x) \cdot log_3(x) \cdots log_{n-1}(x)}\)

Now we need to evaluate the expression:

\(x \cdot log(x) \cdot log_2(x) \cdot log_3(x) \cdots log_{n-1}(x) \cdot log_n(x) \cdot \frac{dy}{dx}\)

Substituting the expression for \(\frac{dy}{dx}\), we get:

\(x \cdot log(x) \cdot log_2(x) \cdot log_3(x) \cdots log_{n-1}(x) \cdot log_n(x) \cdot \frac{1}{x \cdot log(x) \cdot log_2(x) \cdot log_3(x) \cdots log_{n-1}(x)}\)

Simplifying, we get:

\(log_n(x)\)