Question

$A:$ If $x=ct,y=\frac{c}{t}$, then at $t=1,\frac{dy}{dx}=$
$B:$ If $x=3\cos \theta -{\cos }^3\theta ,y=3\sin \theta -{\sin }^3\theta$, then at $\theta =\frac{\pi }{3},\frac{dy}{dx}=$
$C:$ If $x=a(t+\frac{1}{t}),y=a(t-\frac{1}{t})$, then at $t=2,\frac{dy}{dx}=$
$D:$ Derivative of $\log \left(\sec x\right)$ with respect to $\tan x$ at $x=\frac{\pi }{4}$ is
Arrangement of the above values in the increasing order is

• A. $C,D,A,B$
• B. $C,A,D,B$
• C. $A,B,D,C$
• D. $B,D,A,C$

Answer 1

Answer: (C)

$A,B,D,C$

For $A$
$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=-\frac{c}{t^2}\times \frac{1}{c}=-\frac{1}{t^2}$

At $t=1,\frac{dy}{dx}=-1$

For $B$

$\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}$

$=3\cos \theta -3{\sin }^2\theta \times \cos \theta \times \frac{1}{-3\sin \theta +3\sin \theta \times {\cos }^2\theta }$

$=-\frac{3\cos \theta (1-{\sin }^2\theta )}{3\sin \theta (1-{\cos }^2\theta )}=-\frac{{\cos }^3\theta }{{\sin }^3\theta }$

At $\theta =\frac{\pi }{3},\frac{dy}{dx}=-\frac{1}{3\sqrt{3}}$
For $C$
$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}=a(1+\frac{1}{t^2})\times \frac{1}{a(1-\frac{1}{t^2})}=\frac{t^2+1}{t^2-1}$

At $t=2,\frac{dy}{dx}=\frac{5}{3}$
For $D$,
$\frac{dy}{dx}=\frac{\sec x\tan x}{\sec x}\times \frac{1}{{\sec }^{2}x}=\sin x\cos x$

At $x=\frac{\pi }{4},\frac{dy}{dx}=\frac{1}{2}$
So, the arrangement would be $A,B,D,C$