Question

Derivative of $log_2(logx)$ with respect to $x$ is

• A. $\frac{2}{xlogx}$
• B. $\frac{1}{xlogx}$
• C. $\frac{1}{xlogx^2}$
• D. $\frac{2}{logx}$

Answer 1

Answer: (B)

$\frac{1}{xlogx}$

Answer 2

To find the derivative of $log_2(logx)$ with respect to $x$, we use the chain rule and the change of base formula for logarithms.

Let the given function be $y = log_2(logx)$.

First, recall the formula for the derivative of a logarithm with an arbitrary base: $\frac{d}{du}(log_b u) = \frac{1}{u \cdot ln b}$

In our case, the outer function is $log_2(u)$, where $u = logx$. So, applying the chain rule, we have: $\frac{dy}{dx} = \frac{d}{du}(log_2 u) \cdot \frac{du}{dx}$ $\frac{dy}{dx} = \frac{1}{u \cdot ln2} \cdot \frac{d}{dx}(logx)$ Substitute $u = logx$: $\frac{dy}{dx} = \frac{1}{logx \cdot ln2} \cdot \frac{d}{dx}(logx)$

Now, we need to determine the base of the inner logarithm, $logx$. In calculus, when the base of $logx$ is not explicitly specified, it is conventionally understood to be the natural logarithm (base $e$), i.e., $logx = lnx$.

Assuming $logx = lnx$: The derivative of $lnx$ is $\frac{d}{dx}(lnx) = \frac{1}{x}$.

Substitute this into our expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{lnx \cdot ln2} \cdot \frac{1}{x}$ $\frac{dy}{dx} = \frac{1}{x \cdot lnx \cdot ln2}$

Now, let's compare this result with the given options. The options are: A: $\frac{2}{xlogx}$ B: $\frac{1}{xlogx}$ C: $\frac{1}{xlogx^2}$ D: $\frac{2}{logx}$

None of the options contain $ln2$ in the denominator. This suggests a potential common convention or a typo in the question. In many competitive exams, if $logx$ is written without a base, it's often interpreted as $lnx$. If a question involving $log_b(logx)$ results in options without $lnb$, it often implies that the outermost logarithm was also intended to be base $e$.

Let's consider the possibility that the question implicitly intended to ask for the derivative of $ln(logx)$, where $logx$ means $lnx$. If the function was $y = ln(lnx)$: $\frac{dy}{dx} = \frac{1}{lnx} \cdot \frac{d}{dx}(lnx)$ (by chain rule) $\frac{dy}{dx} = \frac{1}{lnx} \cdot \frac{1}{x}$ $\frac{dy}{dx} = \frac{1}{x \cdot lnx}$

If we assume that $logx$ in the options also refers to $lnx$, then $\frac{1}{x \cdot lnx}$ is equivalent to $\frac{1}{x \cdot logx}$. This matches option B.

Given that this is a multiple-choice question and option B is a direct match under the assumption that the outermost logarithm was $ln$ instead of $log_2$, this is the most probable intended answer. The explicit $log_2$ might be a distractor or a mistake in the question phrasing, as the $ln2$ factor is missing from all options.