f'(x)= 1/3x3+ 4x ???
how do i solve this?
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 ^
\(f'(x)= \frac{1}{3}x^3+ 4x \\ f(x)=\int\;\frac{1}{3}x^3+ 4x\;dx \\ f(x)=\frac{1}{3*4}x^4+ \frac{4x^2}{2}+c\\ f(x)=\frac{x^4}{12}+ 2x^2+c\\\)