y=(2×³−7)^4

[(2x)

^{2}-7)]^{4}Rewrite as (-7 + 4x

^{2})^{4}Let u = 4x

^{2}- 7[d/dx] [(4x

^{2}- 7)]^{4}= [du^{4}/du]*[du/dx] where u = 4x^{2}- 7 and [d/du](u^{4}) = 4u^{3}so that

[d/dx] [(4x

^{2}- 7)]^{4}= 4(-7 + 4x^{2})^{3}{[d/dx](-7 + 4x^{2})}^{2}- 7)]

^{4}= 4(-7 + 4x

^{2})

^{3}(0 + 8x) = (8x)(4)(-7 + 4x

^{2})

^{3}[d/dx] [(4x

^{2}- 7)]

^{4}=

**(32x)(-7 + 4x**

^{2})^{3}__

## Comments

^{x}^{3 }or do you mean 2x^{3}? (where only the 3 is the exponent).^{ }