If $\sin x = \frac{2t}{1+t^2}$ and $\tan y = \frac{2t}{1-t^2}$,then $\frac{dy}{dx}$ is equal to

Find the derivative of $f(\sin x)$ with respect to $g(\sec x)$ at $x = \frac{\pi}{4}$, given that $f'(\frac{1}{\sqrt{2}}) = 3$ and $g'(\sqrt{2}) = 1$.

Derivative of $\log (\sec \theta+\tan \theta)$ with respect to $\sec \theta$ at $\theta=\frac{\pi}{4}$ is

If $x=\operatorname{cosec} \theta-\sin \theta$, $y=\operatorname{cosec}^{2022} \theta-\sin ^{2022} \theta$ and $\left(\frac{d y}{d x}\right)^2=\frac{k\left(y^2+4\right)}{g(x)}$ where $k \in R$, then $10+k-g(2022)=$

If $x=f(\theta)$ and $y=g(\theta)$,then $\frac{d^2 y}{d x^2}=$

If $x=e^{t}(\sin t-\cos t)$ and $y=e^{t}(\sin t+\cos t)$,then $\frac{dy}{dx}$ at $t=\frac{\pi}{3}$ is