Indian Navy Pilot Differentiation — practice questions

Explore 2 free Differentiation questions in Indian Navy Pilot Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Differentiation questions and exercises with answers

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$\dfrac{d}{dx}\left[\log(\log x)\right]$ at $x = e^e$ equals:Read the answer and worked solution →If $y = (\sin x)^x$, then $\dfrac{dy}{dx}$ equals:Read the answer and worked solution →

Worked practice questions

$\dfrac{d}{dx}\left[\log(\log x)\right]$ at $x = e^e$ equals:

$\dfrac{1}{(\log x) \cdot x}$ evaluated; = $1/e^{e+1}$
$\dfrac{1}{e \cdot e^e}$
$\dfrac{1}{e^{e+1}}$
$\dfrac{1}{e^e}$

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Correct answer: A. $\dfrac{1}{(\log x) \cdot x}$ evaluated; = $1/e^{e+1}$

$d/dx[\log(\log x)] = (1/\log x) \cdot (1/x) = 1/(x \log x)$. At $x = e^e$: $\log x = e$. **So derivative = $1/(e^e \cdot e) = 1/e^{e+1}$.**

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If $y = (\sin x)^x$, then $\dfrac{dy}{dx}$ equals:

$(\sin x)^x [\sin x + x\cos x]$
$x (\sin x)^{x-1} \cos x$
$(\sin x)^x \cdot \cos x$
$(\sin x)^x [\log(\sin x) + x \cot x]$

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Correct answer: D. $(\sin x)^x [\log(\sin x) + x \cot x]$

**Logarithmic diff:** $\ln y = x \ln(\sin x)$. **Differentiate:** $y'/y = \ln(\sin x) + x \cdot (\cos x/\sin x) = \ln(\sin x) + x\cot x$. **So $y' = y[\ln(\sin x) + x\cot x]$.**