The interval in which y = ln(ln(x)), x 1 is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given function is $y = \ln(\ln(x))$ for $x > 1$. First,we find the derivative of $y$ with respect to $x$: $\frac{dy}{dx} = \frac{1}{\ln(x)} \cdot

Vedclass · Accepted Answer

None of the above

Vedclass · Answer

(D) Given function is $y = \ln(\ln(x))$ for $x > 1$. First,we find the derivative of $y$ with respect to $x$: $\frac{dy}{dx} = \frac{1}{\ln(x)} \cdot \frac{d}{dx}(\ln(x)) = \frac{1}{x \ln(x)}$. For the function to be decreasing,we must have $\frac{dy}{dx} So,$\frac{1}{x \ln(x)} Since the domain is given as $x > 1$,we know that $\ln(x) > 0$ and $x > 1$. Therefore,the product $x \ln(x)$ is always positive for all $x > 1$. Since $\frac{1}{x \ln(x)}$ is always positive for $x > 1$,there is no interval where the function is decreasing. Thus,the correct option is $D$.

The interval in which $y = \ln(\ln(x)), x > 1$ is decreasing is

Similar Questions

If $x \in (0, 1)$,then what type of function is $f(x) = x^{100} + \sin x - 1$?

$A$ real valued function $f:[4, \infty) \rightarrow R$ is defined as $f(x)=(x^2+x+1)^{(x^2-3x-4)}$, then $f$ is

The function $f(x) = 2{x^3} + 18{x^2} - 96x + 45$ is an increasing function when:

If $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$ is a strictly increasing function,then .......

Find the intervals in which the function $f$ given by $f(x)=2x^{3}-3x^{2}-36x+7$ is
$(a)$ increasing
$(b)$ decreasing

Vedclass Test Series

Exam Paper Generator

Online Exam Module