$F\prime \left(t\right)$

$\=f_x x\prime \left(t\right)\+f_y y\prime \left(t\right)$

$\=-\frac{2xt}{\sqrt{1-x^2}\left(2y\+\sqrt{-x^2\+1}\right)}\+\frac{2\cos \left(t\right)}{2y\+\sqrt{1-x^2}}$

$\=-\frac{2t^3}{\sqrt{1-t^4}\left(2\sin \left(t\right)\+\sqrt{-t^4\+1}\right)}\+\frac{2\cos \left(t\right)}{2\sin \left(t\right)\+\sqrt{1-t^4}}$

$\=\frac{2\left(\cos \left(t\right)\sqrt{1-t^4}-t^3\right)}{1-t^4\+2 \sqrt{1-t^4}\sin \left(t\right)}$

Formal statement of the relevant chain rule

$f\left(x\left(t\right)\,y\left(t\right)\right)$$\stackrel{\text{differentiate w.r.t. t}}{\to }$${\mathrm{D}}_1\left(f\right)\left(x\left(t\right)\,y\left(t\right)\right)\left(\frac{\ⅆ}{\ⅆt}x\left(t\right)\right)\+{\mathrm{D}}_2\left(f\right)\left(x\left(t\right)\,y\left(t\right)\right)\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)$$$

Implement the chain rule

$f\left(x\,y\right)\=\ln \left(2y\+\sqrt{1-x^2}\right)$$\stackrel{\text{assign as function}}{\to }$$f$

$X\=t^2$$\stackrel{\text{assign}}{\to }$

$Y\=\sin \left(t\right)$$\stackrel{\text{assign}}{\to }$

$\left(\frac{\partial }{\partial x} f\left(x\,y\right)\right) \left(\frac{\ⅆ}{\ⅆ t} X\right)\+\left(\frac{\partial }{\partial y} f\left(x\,y\right)\right) \left(\frac{\ⅆ}{\ⅆ t} Y\right)$

$-\frac{2xt}{\sqrt{-x^2\+1}\left(2y\+\sqrt{-x^2\+1}\right)}\+\frac{2\cos \left(t\right)}{2y\+\sqrt{-x^2\+1}}$

$\stackrel{\text{evaluate at point}}{\to }$

$-\frac{2t^3}{\sqrt{-t^4\+1}\left(2\sin \left(t\right)\+\sqrt{-t^4\+1}\right)}\+\frac{2\cos \left(t\right)}{2\sin \left(t\right)\+\sqrt{-t^4\+1}}$

$\stackrel{\text{simplify}}{\=}$

$\frac{2\left(-t^3\+\cos \left(t\right)\sqrt{-t^4\+1}\right)}{-t^4\+1\+2\sqrt{-t^4\+1}\sin \left(t\right)}$

Obtain $F\prime \left(t\right)$ from the explicit representation $F\left(t\right)\=f\left(x\left(t\right)\,y\left(t\right)\right)$ 

$\frac{\ⅆ}{\ⅆ t} f\left(X\,Y\right)$ = $\frac{2\cos \left(t\right)-\frac{2t^3}{\sqrt{-t^4\+1}}}{2\sin \left(t\right)\+\sqrt{-t^4\+1}}$$\stackrel{\text{simplify}}{\=}$$\frac{2\left(-t^3\+\cos \left(t\right)\sqrt{-t^4\+1}\right)}{-t^4\+1\+2\sqrt{-t^4\+1}\sin \left(t\right)}$$$

Initialize

$$\begin{gathered} \mathrm{interface}\left(\mathrm{typesetting}\=\mathrm{extended}\right)\: \\ \mathrm{Typesetting}:-\mathrm{Settings}\left(\mathrm{useprime}\,\mathrm{prime}\=t\right)\: \\ \mathrm{Typesetting}:-\mathrm{Suppress}\left(\left[x\left(t\right)\,y\left(t\right)\right]\right)\: \end{gathered}$$

Formal statement of the relevant chain rule

$\mathrm{diff}\left(f\left(x\left(t\right)\,y\left(t\right)\right)\,t\right)$

${\mathrm{D}}_1\left(f\right)\left(x\,y\right)x\prime +{\mathrm{D}}_2\left(f\right)\left(x\,y\right)y\prime$

Implement the chain rule

$\mathrm{Typesetting}:-\mathrm{Unsuppress}\left(\left[x\left(t\right)\,y\left(t\right)\right]\right)\:$

$f≔\left(x\,y\right)\to \ln \left(2y\+\sqrt{1-x^2}\right)\:$

$$\begin{gathered} X≔t^2\: \\ Y≔\sin \left(t\right)\: \end{gathered}$$$$

$\mathrm{simplify}\left({\mathrm{D}}_1\left(f\right)\left(X\,Y\right) \mathrm{diff}\left(X\,t\right)\+{\mathrm{D}}_2\left(f\right)\left(X\,Y\right) \mathrm{diff}\left(Y\,t\right)\right)$

$\frac{2\left(-t^3\+\cos \left(t\right)\sqrt{-t^4\+1}\right)}{-t^4\+1\+2\sqrt{-t^4\+1}\sin \left(t\right)}$

Obtain $F\prime \left(t\right)$ from an explicit representation of $F\left(t\right)$

$\mathrm{simplify}\left(\mathrm{diff}\left(f\left(X\,Y\right)\,t\right)\right)$

$\frac{2\left(-t^3\+\cos \left(t\right)\sqrt{-t^4\+1}\right)}{-t^4\+1\+2\sqrt{-t^4\+1}\sin \left(t\right)}$