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Differential Calculus

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Question : 10 of 69

Marks: +1, -0

If

y=\log (x e^{x})

\;\frac{d y}{d x}

is

$\;\frac{1}{x e^{x}}$
$1+e^{x}$
$1+\;\frac{1}{x}$
none

Solution:

Concept:

Simplify the function using logarithm properties before differentiating.

Explanation:

Given

y = \log(x e^{x})

.

Using

\log(ab) = \log a + \log b

, we get

y = \log x + \log e^{x}

.

Since

\log e^{x} = x

, we have

y = \log x + x

.

Differentiate:

\frac{dy}{dx} = \frac{1}{x} + 1

.

This simplifies to

1 + \frac{1}{x}

.

Answer:

1 + \frac{1}{x}

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