Derivative of Functions in Parametric Forms

$\frac{d^{2}x}{d{y}^2}=$

Differentiate $\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}$ w .r. t. $\sqrt{1-x^4}$

If  $x=a\left(\cos t+t\sin t\right)$ and  $y=a\left(\sin t-t\cos t\right), 0<t<\frac{\pi }{2}.$ The value of   $\frac{d^{2}y}{d{x}^2}$ would be:

If $x={\cos }^{-1}\left(\frac{2u}{1+u^2}\right)$ and $y={\sin }^{-1}\left(\frac{1-u^2}{1+u^2}\right)$, then $\frac{dy}{dx}=$

If the parametric equation of curve is given by $x=\cos \theta +\mathrm{logtan}\frac{\theta }{2}$ and $y=\sin \theta ,$ then the points for which $\frac{dy}{dx}=0$ are given by

Derivative of ${\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$ with respect to ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$ in $\left(1,\infty \right)$, is

The derivative of $a^{\sec x}$ w.r.t. $a^{\tan x} \left(\forall a>0\right)$ is

If $x={\cos }^3\theta$ and $y={\sin }^3\theta$, then $1+{\left(\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}\right)}^2$ is equal to

The derivative of ${\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}\left(\frac{\mathrm{s}\mathrm{i}\mathrm{n}x-\mathrm{c}\mathrm{o}\mathrm{s}x}{\mathrm{s}\mathrm{i}\mathrm{n}x+\mathrm{c}\mathrm{o}\mathrm{s}x}\right)$ with respect to $\frac{x}{2},$ where $x\in \left(0,\frac{\pi }{2}\right)$, is

The derivative of $e^{x^3}$with respect to${\log }_{e}x$is

Given $x=t^3-3t^2+e^t, y=\cos \left(\pi t\right)+ln\left(t^2+1\right),$ then $\frac{dy}{dx}$ at $t=0$

Derivative of $\left(\frac{{\cot }^{-1}\mathrm{x}}{1+{\cot }^{-1}\mathrm{x}}\right)$ with respect to ${\cot }^{-1}\mathrm{x}$ is

If $x=\sin t, y=\cos 2t$, then $\frac{dy}{dx}$ will be

If $x=a{\cos }^{3}t,y=a{\sin }^{3}t$, then value of $\frac{dy}{dx}$ is

Find the derivative of $\log x$ with respect to ${\tan }^{-1}x$

If $x=a(\theta +\sin \theta ),y=a(1-\cos \theta )$, find $\frac{dy}{dx}$.

If $\tan u=\sqrt{\frac{1-x}{1+x}}, \cos v=4x^3-3x$, then $\frac{du}{dv}=$

If $x=3\cos t$ and $y=4\sin t,$ then $\frac{d^{2}y}{d{x}^2}$ at the point $\left(x_0,y_0\right)=\left(\frac{3}{2}\sqrt{2},2\sqrt{2}\right)$ is

The derivative of $f\left(x\right)=x^{{\tan }^{-1}x}$ with respect to $g\left(x\right)={\sec }^{-1}\left(\frac{1}{2x^2-1}\right)$ is

If $y=f\left(x^3\right), z=g\left(x^5\right), f^{\text{'}}\left(x\right)=\tan x$ and $g^{\text{'}}\left(x\right)=\sec x,$ then $\frac{dy}{dz}=$