A) \[\frac{16}{3}\]
B) \[\frac{32}{3}\]
C) \[\frac{16\sqrt{2}}{3}\]
D) \[\frac{32\sqrt{2}}{3}\]
E) \[\frac{32{{(2)}^{1/3}}}{3}\]
Correct Answer: B
Solution :
\[f({{x}^{3}})=4{{x}^{4}}\forall x>0\] Let\[{{x}^{3}}=t\Rightarrow x={{t}^{1/3}}\] \[\therefore \] \[f(t)=4{{t}^{4/3}}\] On differentiating w.r.t. t, we get \[f(t)=4.\frac{4}{3}{{(t)}^{\frac{4}{3}-1}}=4.\frac{4}{3}{{(t)}^{1/3}}\] \[\therefore \] \[f({{x}^{3}})=\frac{16}{3}{{({{x}^{3}})}^{1/3}}=\frac{16}{3}x\] \[\therefore \] \[f(8)=f({{2}^{3}})=\frac{16}{3}\times 2=\frac{32}{3}\]
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