If $x=e^\theta(\sin \theta-\cos \theta)$ and $y=e^\theta(\sin \theta+\cos \theta)$,then $\frac{dy}{dx}$ at $\theta=\frac{\pi}{4}$ is:

$A$ curve is given by the equations $x = a \cos \theta + \frac{1}{2}b \cos 2\theta$ and $y = a \sin \theta + \frac{1}{2}b \sin 2\theta$. The points for which $\frac{d^2y}{dx^2} = 0$ are given by:

For $a > 0, t \in \left( 0, \frac{\pi}{2} \right)$,let $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$. Then,$1 + \left( \frac{dy}{dx} \right)^2$ equals

The position of a point at time $t$ is given by $x = a + bt - ct^2$ and $y = at + bt^2$. Its acceleration at time $t$ is:

The function $y(x)$ represented by $x=\sin t$,$y=a e^{t \sqrt{2}}+b e^{-t \sqrt{2}}$,$t \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ satisfies the equation $(1-x^2) y^{\prime \prime}-x y^{\prime}=k y$,then the value of $k$ is

The $2^{nd}$ derivative of $a \sin^3 t$ with respect to $a \cos^3 t$ at $t = \frac{\pi}{4}$ is