MCQ
If $y =\sqrt[3]{\left(3 x^2+8 x-6\right)^5}$, then $\frac{ d y}{ d x}=$ ?
- ✓$\frac{5}{3}(6 x+8)\left(3 x^2+8 x-6\right)^{\frac{2}{3}}$
- B$\frac{-5}{3}(6 x+8)\left(3 x^2+8 x-6\right)^{\frac{2}{3}}$
- C$\frac{3}{5}(3 x+4)\left(3 x^2+8 x-6\right)^{\frac{2}{3}}$
- D$\frac{-3}{5}(3 x+4)\left(3 x^2+8 x-6\right)^{\frac{2}{3}}$