Question

Tính đạo hàm hàm số :

\(y=\sqrt[5]{2x^3+1}\)

Answer 1

Ta có : \(y'=\frac{\left(2x^3+1\right)'}{5\sqrt[5]{\left(2x^3+1\right)^4}}=\frac{6x^2}{5\sqrt[5]{\left(2x^3+1\right)^4}}\)

Answer 2

Ta có : 

                \(y'=\frac{\left(2x^3+1\right)'}{5\sqrt[5]{\left(2x^3+1\right)^4}}=\frac{6x^2}{5\sqrt[5]{\left(2x^3+1\right)^4}}\)

Answer 3

\(y'=\frac{\left(2x^3+1\right)}{5\sqrt[5]{\left(2x^3+1\right)^4}}=\frac{6x^2}{5\sqrt[5]{\left(2x^3+1\right)^4}}\)

Answer 4

\(y'=\frac{\left(2x^3+1\right)}{5\sqrt[5]{\left(2x^3+1\right)^4}}=\frac{6x^2}{5\sqrt[5]{\left(2x^3+1\right)^4}}\)

Answer 5

\(y'=\frac{\left(2x^3+1\right)}{5\sqrt[5]{\left(2x^3+1\right)^4}}\)  =\(\frac{6x^2}{5\sqrt[5]{\left(2x^3+1\right)^4}}\)