Question

Find the derivative of the function. f(x)=(1+x4)2/3f(x)=\left(1+x^{4}\right)^{2 / 3}

Solution

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f(x)=(1+x4)2/3Apply The Power Rule Combined with the Chain Ruleddx[g(x)]n=n[g(x)]n1g(x)Therefore,f(x)=23(1+x4)2/31n[g(x)]n1ddx[1+x4]g(x)f(x)=23(1+x4)1/3(4x3)Simplify and rewritef(x)=8x33(1+x4)1/3f(x)=8x331+x43\begin{gathered} f\left( x \right) = {\left( {1 + {x^4}} \right)^{2/3}} \\ \textcolor{#4257b2}{{\text{Apply The Power Rule Combined with the Chain Rule}}} \\ \frac{d}{{dx}}{\left[ {g\left( x \right)} \right]^n} = n{\left[ {g\left( x \right)} \right]^{n - 1}} \cdot g'\left( x \right) \\ \textcolor{#4257b2}{ {\text{Therefore}}{\text{,}}} \\ f'\left( x \right) = \overbrace {\frac{2}{3}{{\left( {1 + {x^4}} \right)}^{2/3 - 1}}}^{n{{\left[ {g\left( x \right)} \right]}^{n - 1}}}\overbrace {\frac{d}{{dx}}\left[ {1 + {x^4}} \right]}^{g'\left( x \right)} \\ f'\left( x \right) = \frac{2}{3}{\left( {1 + {x^4}} \right)^{ - 1/3}}\left( {4{x^3}} \right) \\ \textcolor{#4257b2}{ {\text{Simplify and rewrite}}} \\ f'\left( x \right) = \frac{{8{x^3}}}{{3{{\left( {1 + {x^4}} \right)}^{1/3}}}} \\ f'\left( x \right) = \frac{{8{x^3}}}{{3\sqrt[3]{{1 + {x^4}}}}} \\ \end{gathered}

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