Question

Logarithmic diffrentiation:

Answer 1

\[y=\sqrt{x} ^{x}\]

Answer 2

Taking natural logarithm both sides, we have
\[ \ln(y) =\frac{1}{2}x\ln(x) \]
Differentiate both sides with respect to x, we have
\[ \frac{d\ln(y)}{dx}=\frac{d}{dx}\left(\frac{1}{2}x\ln(x)\right) \]
therefore,
\[ \frac{1}{y}\frac{dy}{dx}= \frac{1}{2}\left(\ln(x)+1\right) \]
or
\[ \frac{dy}{dx}=\frac{\sqrt{x}^x}{2}(\ln(x)+1) \]

Answer 3

\[f(bob) = bobbobbobbob \implies \log f(bob) = 4\log (bob) \] Differentiate it.
\[\frac{1}{f(bob)}(f(bob))' = \frac{4}{bob}(bob)' \implies f'(bob) = f(bob)\cdot\frac4{bob}\cdot (bob)' \]

Answer 4

@ishaan , classic!

Answer 5

lol thanks