MCQ
Differential coefficient of  ${\sin ^{ - 1}}x$ $w.r.t$ ${\cos ^{ - 1}}\sqrt {1 - {x^2}} $ is
  • $1$
  • B
    ${1 \over {1 + {x^2}}}$
  • C
    $2$
  • D
    None of these

Answer

Correct option: A.
$1$
a
(a) Let ${y_1} = {\sin ^{ - 1}}x$ and ${y_2} = {\cos ^{ - 1}}\sqrt {1 - {x^2}} $

Differentiating w.r.t. $x$ of ${y_1}$ and ${y_2}$, we get

$\frac{{d{y_1}}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }}$

$\frac{{d{y_2}}}{{dx}} = - \frac{1}{{\sqrt {1 - (1 - {x^2})} }}\frac{{1( - 2x)}}{{2\sqrt {1 - x} }} $

$= \frac{1}{{\sqrt {1 - {x^2}} }} \Rightarrow \frac{{d{y_2}}}{{d{y_1}}} = 1.$

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

Similar questions

Let $L$ be a normal to the parabola $y^2=4 x$. If $L$ passes through the point $(9,6)$, then $L$ is given by

$(A)$ $y-x+3=0$ $(B)$ $y+3 x-33=0$ $(C)$ $y+x-15=0$ $(D)$ $y-2 x+12=0$

If $a \in R$ and the equation $ - 3{\left( {x - \left[ x \right]} \right)^2} + 2\left( {x - \left[ x \right]} \right) + {a^2} = 0$ (where $[x]$ denotes the greatest integer $\leq\,x$) has no integral solution, then all possible values of $a$ lie in the interval
Locus of the feet of the perpendiculars drawn from either foci on a variable tangent to the hyperbola $16y^2 - 9x^2 = 1 $ is
For any positive integer $n$, let $S_n:(0, \infty) \rightarrow R$ be defined by

$S_n(x)=\sum_{k=1}^n \cot ^{-1}\left(\frac{1+k(k+1) x^2}{x}\right)$

where for any $x \in R , \cot ^{-1} x \in(0, \pi)$ and $\tan ^{-1}(x) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then which of the following

statements is (are) $TRUE$?

$(A)$ $S _{10}( x )=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1+11 x ^2}{10 x }\right)$, for all $x >0$

$(B)$ $\lim _{n \rightarrow \infty} \cot \left(S_n(x)\right)=x$, for all $x>0$

$(C)$ The equation $S_3(x)=\frac{\pi}{4}$ has a root in $(0, \infty)$

$(D)$ $\tan \left( S _{ n }( x )\right) \leq \frac{1}{2}$, for all $n \geq 1$ and $x >0$

$\int_{}^{} {\frac{{1 + \tan x}}{{x + \log \sec x}}\;dx = } $
$\frac{{\sqrt {5 + 12i} + \sqrt {5 - 12i} }}{{\sqrt {5 + 12i} - \sqrt {5 - 12i} }} = $
If $f(x) = {x^2} - 1$ and $g(x) = 3x + 1$, then $(gof)(x) = $
If the three normals drawn to the parabola, $y ^{2}=2 x$ pass through the point $( a , 0) a \neq 0,$ then $'a'$ must be greater than
The area of the region $\left\{ {\left( {x,y} \right):{x^2} + {y^2} \leqslant 1 \leqslant x + y} \right\}$ is
If $y = {\tan ^{ - 1}}\left( {{x \over {1 + \sqrt {1 - {x^2}} }}} \right)$, then ${{dy} \over {dx}} = $