If {f}(x)= ₓ²}( x), then at x=e, {f}^{}(x) has the value

Correct Answer is : 12e1/2 e2e1

Correct Answer is : 12e1/2 e2e1

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MHTCET 11 May 2024 Shift 2 Paper

Question : 114 of 150

Marks: +1, -0

\mathrm{f}(x)=\log _{x^{2}}(\log x)

x=e, \mathrm{f}^{\prime}(x)

has the value

Solution:

Concept:

Convert the logarithm to natural log and differentiate using the quotient rule.

Explanation:

Let

y = f(x) = \log_{x^2}(\log x)

Using the base change formula,

y = \frac{\ln(\ln x)}{\ln(x^2)} = \frac{\ln(\ln x)}{2\ln x}

Differentiate with respect to

x

using the quotient rule:

f'(x) = \frac{\frac{1}{x\ln x} \cdot 2\ln x - \ln(\ln x) \cdot \frac{2}{x}}{(2\ln x)^2}

Simplify the numerator:

f'(x) = \frac{\frac{2}{x} - \frac{2\ln(\ln x)}{x}}{4(\ln x)^2} = \frac{1 - \ln(\ln x)}{2x(\ln x)^2}

Now substitute

x = e

\ln e = 1

and

\ln(\ln e) = \ln 1 = 0

f'(e) = \frac{1 - 0}{2e(1)^2} = \frac{1}{2e}

Answer:

f'(e) = \frac{1}{2e}

So the correct option is D.

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