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Problem :

A function g defined for all real $x>0$ satisfies $g(1)=1, g'(x^2)=x^3$ for all $x>0$ then find g(4)

Please suggest how to tackle this getting no clue how to proceed tried to form function by hit and trial method but not getting the proper function, please help. thanks

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If $x^2=t>0$ then $g'(t)=t^{3/2}.$ Then $g(t)=\frac{2}{5}t^{5/2}+C$. From $g(1)=1$ it follows that $C=\frac{3}{5}.$ Then $g(x^2)=\frac{2}{5}x^5+\frac{3}{5}$ and $g(2^2)=\frac{2}{5}2^5+\frac{3}{5}=\frac{67}{5}.$

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