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f'(x)= 1/3x3+ 4x ???

 

how do i solve this?

 Jul 7, 2017
 #1
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f'(x)= 1/3x3+ 4x 

f'(x)= (1/3)x^3+ 4x 

Is this what you mean?   The symbol for 'to the power of' is ^

 

\(f'(x)= \frac{1}{3}x^3+ 4x \\ f(x)=\int\;\frac{1}{3}x^3+ 4x\;dx \\ f(x)=\frac{1}{3*4}x^4+ \frac{4x^2}{2}+c\\ f(x)=\frac{x^4}{12}+ 2x^2+c\\\)

 Jul 7, 2017

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