differentiate
1⁄3x^6(x^2-3)
y=1/3x6(x2−3)y=x63×(x2−3)y=x83−x6dydx=8x73−6x5
1⁄3x^6(x^2-3) \\y= 1/3x^6(x^2-3)\\\\ y=\frac{x^6}{3}\times (x^2-3)\\\\ y=\frac{x^8}{3}-x^6\\\\ \frac{dy}{dx}=\frac{8x^7}{3}-6x^5\\\\ Like melody's answer