Sprawdźmy A.

$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)=f\left(\sqrt{x}\right)={\sqrt{x}}^3+{\sqrt{x}}^2=x\sqrt{x}+x$  

 

Sprawdźmy B.     

$\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)=g\left(x^3+x^2\right)=\sqrt{x^3+x^2}=\sqrt{x^2\left(x+1\right)}=x\sqrt{x+1}$  

 

Sprawdźmy C.

$\left(f\left(g\left(x\right)\right)\right){}^{\prime }=\left({\sqrt{x}}^3+{\sqrt{x}}^2\right){}^{\prime }=\left(x^{\frac{3}{2}}+x\right){}^{\prime }=\frac{3}{2}{x}^{\frac{3}{2}-1}+1=\frac{3}{2}{x}^{\frac{1}{2}}+1=\frac{3}{2}\sqrt{x}+1$  

 

Sprawdźmy D.

$\left(g\left(f\left(x\right)\right)\right){}^{\prime }=\left(x\sqrt{x+1}\right){}^{\prime }=\left(x\right){}^{\prime }\sqrt{x+1}+x\cdot \left(\sqrt{x+1}\right){}^{\prime }=\sqrt{x+1}+x\cdot \left({\left(x+1\right)}^{\frac{1}{2}}\right){}^{\prime }=$  

$\sqrt{x+1}+x\cdot \frac{1}{2}{\left(x+1\right)}^{\frac{1}{2}-1}=\sqrt{x+1}+\frac{1}{2}x{\left(x+1\right)}^{-\frac{1}{2}}$  

 

Odp. D