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
    $\frac{-3}{5}(3 x+4)\left(3 x^2+8 x-6\right)^{\frac{2}{3}}$

Answer

Correct option: A.
$\frac{5}{3}(6 x+8)\left(3 x^2+8 x-6\right)^{\frac{2}{3}}$
$\frac{5}{3}(6 x+8)\left(3 x^2+8 x-6\right)^{\frac{2}{3}}$

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