Download App

CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY1 Mark
MCQ
If $\text{x}=\text{f}(\text{t})$ and $\text{y}=\text{g}(\text{t}),$ then $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ is equals to:
  • ✓
    $\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$
  • B
    $\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^2}$
  • C
    $\frac{\text{g}''}{\text{f}''}$
  • D
    $\frac{\text{f}''\text{g}'-\text{g}''\text{f}'}{(\text{g}')^3}$
✓

Answer

Correct option: A.
$\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$
$\text{x}=\text{f}(\text{t})$ $\text{y}=\text{g}(\text{t}),$

$\frac{\text{dy}}{\text{dx}}=\frac{\frac{\text{dy}}{\text{dt}}}{\frac{\text{dx}}{\text{dt}}}=\frac{\text{g}'}{\text{f}'}$

$\frac{\text{d}^2\text{y}}{\text{dx}^2}$

$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)$

$=\frac{\text{d}}{\text{dx}}\Big(\frac{\text{dy}}{\text{dx}}\Big)\frac{\text{dt}}{\text{dx}}$

$=\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{\text{f}'^2}\frac{1}{\text{f}'}$

$=\frac{\text{f}'\text{g}''-\text{g}'\text{f}''}{(\text{f}')^3}$

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 CONTINUITY AND DIFFERENTIABILITY→CONTINUITY AND DIFFERENTIABILITY — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

$\text{f}(\text{x})=2\text{x}-\tan^{-1}-\log\Big\{\text{x}+\sqrt{\text{x}^2+1}\Big\}$ is monotonically increasing when:
→
The existence of the unique solution of the system of equations:
$x + y + z = \lambda$
$5x − y + µz = 10$
$2x + 3y − z = 6$
depends on
→
The total number of distinct $x \in[0,1]$ for which $\int_0^x \frac{t^2}{1+t^4} d t=2 x-1$ is
→
The area of the region bounded by the ellipse $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{16}=1$ is:
→
If three points $A, B $ and $ C$ have position vectors $(1,\,x,\,3),\,\,(3,\,4,\,7)$ and $(y,-2-5)$ respectively and if they are collinear, then $(x,\,y) = $
→
The degree of the differntial equation $\left\{5+\Big(\frac{\text{dy}}{\text{dx}}\Big)^{2}\right\}^{\frac{5}{3}}=\text{x}^{5}\Big(\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}\Big)$ is:
→
If $\vec{a}$ and $\vec{b}$ are vectors in space given by $\vec{a}=\frac{\hat{i}-2 \hat{j}}{\sqrt{5}}$ and $\vec{b}=\frac{2 \hat{i}+\hat{j}+3 \hat{k}}{\sqrt{14}}$, then the value of $(2 \vec{a}+\vec{b}) \cdot[(\vec{a} \times \vec{b}) \times(\vec{a}-2 \vec{b})]$ is
→
If the minimum and the maximum values of the function $f :\left[\frac{\pi}{4}, \frac{\pi}{2}\right] \rightarrow R ,$ defined by : 

$f (\theta)=\left|\begin{array}{ccc}-\sin ^{2} \theta & -1-\sin ^{2} \theta & 1 \\ -\cos ^{2} \theta & -1-\cos ^{2} \theta & 1 \\ 12 & 10 & -2\end{array}\right|$ are $m$ and $M$ respectively, then the ordered pair $( m , M )$ is equal to

→
A random variable $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3.$ If $P(X = 3) = 2P(X = 1)$ and $P(X = 2) = 0.3,$ then $P(X = 0)$ is:
→
The correct evaluation of $\int_0^\pi {\left| {\,{{\sin }^4}x\,} \right|\,dx} $ is
→