MCQ
${d \over {dx}}{\cos ^{ - 1}}\sqrt {{{1 + {x^2}} \over 2}} = $
- A${{ - 1} \over {2\sqrt {1 - {x^4}} }}$
- B${1 \over {2\sqrt {1 - {x^4}} }}$
- ✓${{ - x} \over {\sqrt {1 - {x^4}} }}$
- D${x \over {\sqrt {1 - {x^4}} }}$
✓
Answer
Correct option: C.
${{ - x} \over {\sqrt {1 - {x^4}} }}$
c
(c) Putting ${x^2} = \cos 2\theta $, we have
(c) Putting ${x^2} = \cos 2\theta $, we have
$\frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}\sqrt {\frac{{1 + {x^2}}}{2}} } \right] = \frac{d}{{dx}}[{\cos ^{ - 1}}\cos \theta ]$
$ = \frac{d}{{dx}}[\theta ] = \frac{d}{{dx}}\left[ {\frac{1}{2}{{\cos }^{ - 1}}{x^2}} \right] = \frac{{ - x}}{{\sqrt {1 - {x^4}} }}$.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In a triangle $\mathrm{ABC}$, if $|\overrightarrow{\mathrm{BC}}|=3,|\overrightarrow{\mathrm{C}}|=5$ and $|\overrightarrow{\mathrm{BA}}|=7$, then the projection of the vector $\overline{\mathrm{BA}}$ on $\overline{\mathrm{BC}}$ is equal to:
→If $(\vec{a}+3 \vec{b})$ is perpendicular to $(7 \vec{a}-5 \vec{b})$ and $(\vec{a}-4 \vec{b})$ is perpendicular to $(7 \vec{a}-2 \vec{b})$, then the angle between $\vec{a}$ and $\vec{b}$ (in degrees) is $......$
→Which one of the following statements is not true:
The area bounded by the circles $= 4x^2 + y^2 = 1, x^2 + y^2 = 4$ in the first Quadrant is:
→If a curve $y = a\sqrt x + bx$ passes through the point $(1, 2)$ and the area bounded by the curve, line $x = 4$ and $x-$ axis is $8$ sq. unit, then
→$\text{The value of}\ \hat{\text{i}}\cdot(\hat{\text{j}}\times\hat{\text{k}})+\hat{\text{j}}\cdot(\hat{\text{i}}\times\hat{\text{k}})+\hat{\text{k}}\cdot(\hat{\text{i}}\times\hat{\text{j}})\ \text{is}$
→The curve $\text{y}=\text{x}^{\frac{1}{5}}$ has at $(0, 0)$
→The product of the d.cs of the line which makesequal angles with ox, oy, oz is:
The shaded region in the given figure is a graph of $.....$
→The soluton of the differential equation $\frac{{dy}}{{dx}} = \frac{{x + y}}{x}$ satisfying the condition $y\left( 1 \right) = 1\;$
→