Differentiate {-log (log x), x > 1} with respect to x

Question

Question

This question was previously asked in

NIMCET 2017 Official Paper

Answer 1

Concept:

Chain rule:

\(\rm \frac{d}{dx}[f(g(x))]= f'(g(x)) g'(x)\)

Calculation:

Here, -log (log x), x > 1

Let, log x = y

Differentiating with respect to x, we get

⇒ dy/dx = 1/x ....(1)

Now, -log (log x) = -log y

\(\rm \frac{d}{dx}(-\log y)=-\frac{1}{y}\frac{dy}{dx}\\ \)

= -1 / (x log x) ....(from (1))

Hence, option (1) is correct.