MCQ
What is the value of $\frac{\text{d}}{\text{dx}}\text{(ex sinx + ex cos ⁡x)}?$
  • A
    $0$
  • B
    $\text{2 cos⁡x}$
  • C
    $2e^x.\sin ⁡x$
  • $2e^x.\cos⁡ x$

Answer

Correct option: D.
$2e^x.\cos⁡ x$
We need to use product rule in both the terms to get the answer.
$=\frac{d}{dx}(\text{f.g})=\text{g}\frac{d}{dx}(\text{f})+\text{f}\frac{d}{dx}(\text{fg})$
$=\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{⁡x}) = ({\text{e}^{\text{x}},.\frac{\text{d}}{\text{dx}}} (\sin⁡\text{x}) + \text{sin ⁡x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}})) + (\text{e}^{\text{x}}.\frac{\text{d}}{\text{dx}} (\cos \text{⁡x}) + \cos⁡ \text{x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}}))$
$=\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{⁡x}) =(\text{e}^{\text{x}}.\cos⁡ \text{x} + \sin \text{⁡x} . \text{e}^{\text{x}}) + (\text{e}^{\text{x}}.(-\sin ⁡\text{x}) + \cos ⁡\text{x}.\text{e}^{\text{x}})$
$= \frac{\text{d}}{\text{dx}} (\text{e}^{\text {x}} \sin \text{x} + \text{e}^{\text {x}} \cos \text{⁡x}) = \text{e}^{\text {x}}.\cos⁡ \text{x} + \sin⁡ \text{x} . \text{e}^{\text {x}} – \text{ex}.\sin⁡ \text{x} + \cos \text{⁡x}.\text{e}^{\text {x}}$
$= \frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{x}) = 2\text{e}^{\text{x}}.\text{cos}⁡ \text{x}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free