MCQ
What is the value of $\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \tan \text{x})$ at $\text{x} = 0?$
  • A
    $0$
  • $1$
  • C
    $-1$
  • D
    $2$

Answer

Correct option: B.
$1$
We need to use product rule in both the terms to get the answer.
$\frac{\text{d}}{\text{dx}} (\text{f.g}) = \text{g}.\frac{\text{d}}{\text{dx}} (\text{f})+ (\text{f}).\frac{\text{dy}}{\text{dx}} (\text{g})$
Here $f = e^x$ and $g = \tan ⁡x$
$ \frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \tan \text{x}) = \tan⁡ \text{x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}}) + \text{e}^{\text{x}}.\frac{\text{d}}{\text{dx}}(\tan ⁡\text{x})$
$\frac{d}{dx} (\text{e}^{\text{x}} \tan \text{x}) = \tan⁡ \text{e}^{\text{x}} + \text{e}^{\text{x}}.\sec2\text {⁡x}$
At $x = 0$ we get,
$= \tan ⁡0.\text{e}0 + \text{e}0.\sec2⁡0$
$= 0.(1) + 1.(1)$
$= 1$

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