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

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

$\=-\frac{1}{2}\frac{{\ⅇ}^{2y}\cos \left(4z\right)}{\sqrt{1-t}}\+2x{\ⅇ}^{2y}\cos \left(4z\right)\+8x{\ⅇ}^{2y}\sin \left(4z\right)t$

$\=-\frac{1}{2}\frac{{\ⅇ}^{2t}\cos \left(4t^2-4\right)}{\sqrt{1-t}}\+2\sqrt{1-t}{\ⅇ}^{2t}\cos \left(4t^2-4\right)-8\sqrt{1-t}{\ⅇ}^{2t}t\sin \left(4t^2-4\right)$

Formal statement of the relevant chain rule

$f\left(x\left(t\right)\,y\left(t\right)\,z\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)\,z\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)\,z\left(t\right)\right)\left(\frac{\ⅆ}{\ⅆt}y\left(t\right)\right)\+{\mathrm{D}}_3\left(f\right)\left(x\left(t\right)\,y\left(t\right)\,z\left(t\right)\right)\left(\frac{\ⅆ}{\ⅆt}z\left(t\right)\right)$

Implement the chain rule

$f\left(x\,y\,z\right)\=x{\ⅇ}^{2y}\cos \left(4z\right)$$\stackrel{\text{assign as function}}{\to }$$f$

$X\=\sqrt{1-t}$$\stackrel{\text{assign}}{\to }$$$

$Y\=t$$\stackrel{\text{assign}}{\to }$

$Z\=1-t^2$$\stackrel{\text{assign}}{\to }$

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

$-\frac{1}{2}\frac{{\ⅇ}^{2y}\cos \left(4z\right)}{\sqrt{1-t}}\+2x{\ⅇ}^{2y}\cos \left(4z\right)\+8x{\ⅇ}^{2y}\sin \left(4z\right)t$

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

$-\frac{1}{2}\frac{{\ⅇ}^{2t}\cos \left(4t^2-4\right)}{\sqrt{1-t}}\+2\sqrt{1-t}{\ⅇ}^{2t}\cos \left(4t^2-4\right)-8\sqrt{1-t}{\ⅇ}^{2t}t\sin \left(4t^2-4\right)$

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

$\frac{\ⅆ}{\ⅆ t} f\left(X\,Y\,Z\right)$ = $-\frac{1}{2}\frac{{\ⅇ}^{2t}\cos \left(4t^2-4\right)}{\sqrt{1-t}}\+2\sqrt{1-t}{\ⅇ}^{2t}\cos \left(4t^2-4\right)-8\sqrt{1-t}{\ⅇ}^{2t}t\sin \left(4t^2-4\right)$

Initialize

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

Formal statement of the relevant chain rule

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

${\mathrm{D}}_1\left(f\right)\left(x\,y\,z\right)x\prime +{\mathrm{D}}_2\left(f\right)\left(x\,y\,z\right)y\prime +{\mathrm{D}}_3\left(f\right)\left(x\,y\,z\right)z\prime$

Implement the chain rule

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

$f≔\left(x\,y\,z\right)\to x{\ⅇ}^{2y}\cos \left(4z\right)\:$

$$\begin{gathered} X≔\sqrt{1-t}\: \\ Y≔t\: \\ Z≔1-t^2 \end{gathered}$$

${\mathrm{D}}_1\left(f\right)\left(X\,Y\,Z\right) \mathrm{diff}\left(X\,t\right)\+{\mathrm{D}}_2\left(f\right)\left(X\,Y\,Z\right) \mathrm{diff}\left(Y\,t\right)\+{\mathrm{D}}_3\left(f\right)\left(X\,Y\,Z\right) \mathrm{diff}\left(Z\,t\right)$

$-\frac{{\ⅇ}^{2t}\cos \left(4t^2-4\right)}{2\sqrt{1-t}}+2\sqrt{1-t}{\ⅇ}^{2t}\cos \left(4t^2-4\right)-8\sqrt{1-t}{\ⅇ}^{2t}\sin \left(4t^2-4\right)t$

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

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

$-\frac{{\ⅇ}^{2t}\cos \left(4t^2-4\right)}{2\sqrt{1-t}}+2\sqrt{1-t}{\ⅇ}^{2t}\cos \left(4t^2-4\right)-8\sqrt{1-t}{\ⅇ}^{2t}\sin \left(4t^2-4\right)t$