MCQ
What is the value of ddx (sin x3 cos x2)?
- A
- B3x2cos 2 cos x3-2x sin x3 sin x2
- C2x cos x2 cos x3-2x sin x3 sin x2
- D
Solution:
We follow product rule $\frac{\text{d}}{\text{dx}}(\text{f}.\text{g}.)=\text{g}.\frac{\text{d}}{\text{dx}}(\text{f})+\text{f}.\frac{\text{dy}}{\text{dx}}(\text{g})$
Here $ \text{f} = \sin \text{x}3 \text{ and g} = \cos\text{x}2$
$\frac{\text{d}}{\text{dx}} (\text{f}) = \text{3x2} \text{ cos x3}$
$\frac{\text{d}}{\text{dx}} (\text{g}) = -2\text{x}\ \sin \text{x}^2$
We now substitute this in our main equation,
$= \text{cos x2 .3x2 cos x3 + sin x3.(-2x sin x2)}$
$=\text{ 3x2 cos x2 cos x3 – 2x sin x3 sin x2}$
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Let Sn denote the sum of the cubes of the first n natural numbers and sn denote the sum of the first n natural numbers. Then $\sum\limits^\text{n}_{\text{r}=1}\frac{\text{S}_\text{r}}{\text{S}_\text{r}}$ equals to:
$\frac{\text{n}(\text{n}+1)(\text{n}+2)}{6}$
$\frac{\text{n}(\text{n}+1)}{2}$
$\frac{\text{n}^{2}+3\text{n}+2}{2}$
None of these.