MCQ
What is the value of $\frac{\text{d}}{\text{dx}}\text{(ex sinx + ex cos x)}?$
- A$0$
- B$\text{2 cosx}$
- C$\text{2ex.sin x}$
- D$\text{2ex.cos x}$
Solution:
We need to use product rule in both the terms to get the answer.
$=\frac{d}{dx}(\text{f.g})=\text{g}\frac{d}{dx}(\text{f})+\text{f}\frac{d}{dx}(\text{fg})$
$=\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{x}) = ({\text{e}^{\text{x}},.\frac{\text{d}}{\text{dx}}} (\sin\text{x}) \\+ \text{sin x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}})) + (\text{e}^{\text{x}}.\frac{\text{d}}{\text{dx}} (\cos \text{x}) + \cos \text{x}.\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}}))$
$=\frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{x}) =(\text{e}^{\text{x}}.\cos \text{x} + \sin \text{x} . \text{e}^{\text{x}}) + (\text{e}^{\text{x}}.(-\sin \text{x}) + \cos \text{x}.\text{e}^{\text{x}})$
$= \frac{\text{d}}{\text{dx}} (\text{e}^{\text {x}} \sin \text{x} + \text{e}^{\text {x}} \cos \text{x}) = \text{e}^{\text {x}}.\cos \text{x} + \sin \text{x} . \text{e}^{\text {x}} – \text{ex}.\sin \text{x} + \cos \text{x}.\text{e}^{\text {x}}$
$= \frac{\text{d}}{\text{dx}} (\text{e}^{\text{x}} \sin \text{x} + \text{e}^{\text{x}} \cos \text{x}) = 2\text{e}^{\text{x}}.\text{cos} \text{x}$
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If $2\tan\alpha=3\tan\beta$ then $\tan(\alpha-\beta)=$
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