Download App

STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
MCQ
If $u = {\sin ^{ - 1}}\left( {{y \over x}} \right),$ then ${{\partial u} \over {\partial x}}$ is equal to
  • A
    $ - {y \over {{x^2} + {y^2}}}$
  • B
    ${x \over {\sqrt {1 - {y^2}} }}$
  • C
    ${{ - y} \over {\sqrt {{x^2} - {y^2}} }}$
  • ✓
    ${{ - y} \over {x\sqrt {{x^2} - {y^2}} }}$
✓

Answer

Correct option: D.
${{ - y} \over {x\sqrt {{x^2} - {y^2}} }}$
d
(d) $u = {\sin ^{ - 1}}\frac{y}{x}$;

$\therefore $ $\frac{{\partial u}}{{\partial x}} = \frac{1}{{\sqrt {1 - \frac{{{y^2}}}{{{x^2}}}} }}.\left( { - \frac{y}{{{x^2}}}} \right) = - \frac{y}{{x\sqrt {{x^2} - {y^2}} }}$.

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

Explore more

More "M.C.Q (1 Marks)" from STD 12 - 5. continuity and differentiation→STD 12 - 5. continuity and differentiation — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

Let $f(x) = \int {\frac{{{x^2}dx}}{{(1 + {x^2})\,\left( {1 + \sqrt {1 + {x^2}} } \right)}}} $and $f(0) = 0$, then the value of $f(1)$ be
→
Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y)$ for all $x, y \in R$, and $g: R \rightarrow(0, \infty)$ be a function such that $g(x+y)=g(x) g(y)$ for all $x, y \in R$. If $f\left(\frac{-3}{5}\right)=12$ and $g\left(\frac{-1}{3}\right)=2$, then the value of $\left(f\left(\frac{1}{4}\right)+g(-2)-8\right) g(0)$ is. . . . . 
→
The value of $k$ for which the function $f\left( x \right) = \left\{ \begin{gathered} {\left( {\frac{4}{5}} \right)^{\frac{{\tan \,4x}}{{\tan \,5x}}}},\,\,\,\,0 < x < \frac{\pi }{2} \hfill \\  k + \frac{2}{5}\,\,\,,\,\,\,\,\,\,\,\,\,\,\,x = \frac{\pi }{2} \hfill \\ \end{gathered}  \right.$ is continuous at  $x\,= \frac{\pi}{2}$ is
→
The value of $\int_{\,0}^{\,1} {\,|\,3{x^2} - 1\,|\,dx} $ is
→
Let $f, g : R \rightarrow R$ be defined by $f(x) = 3x + 1$ and $g(x) = x^2 - 2, \forall x \in R,$ respectively. Then, $fog$ is:
→
Consider a curve $y=y(x)$ in the first quadrant as shown in the figure. Let the area $A_{1}$ is twice the area $A _{2}$. Then the normal to the curve perpendicular to the line $2 x -12 y =15$ does NOT pass through the point.
→
The function $\text{f(x)}=\frac{\text{x}^3+\text{x}^2-16\text{x}+20}{\text{x}-2}$ is not defind for $x = 2$. in order to make $f(x)$ continuous at $x = 2$, here $f(2)$ should be defined as :
→
The value of f(0), so that the function $\text{f(x)}=\frac{2-(256-7\text{x})^{\frac{1}{8}}}{(5\text{x}+32)^\frac{1}{5}-2},\text{ x}\neq0$ is continuous everywhere, is given by:
→
Let a unit vector which makes an angle of $60^{\circ}$ with $2 \hat{i}+2 \hat{j}-\hat{k}$ and an angle of $45^{\circ}$ with $\hat{i}-\hat{k}$ be $\overrightarrow{\mathrm{C}}$. Then $\overrightarrow{\mathrm{C}}+\left(-\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3 \sqrt{2}} \hat{\mathrm{j}}-\frac{\sqrt{2}}{3} \hat{\mathrm{k}}\right)$ is :
→
The value of integral $\int_{1/\pi }^{2/\pi } {\frac{{\sin (1/x)}}{{{x^2}}}} \,dx = $
→