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How do you differentiate $3 {(x^{2} - 2)}^{4}$ $3(x^2-2)^4$ ?

1 Answer

Apr 6, 2018

$\frac{d z}{d x} = 24 x {(x^{2} - 2)}^{3}$ $(dz)/(dx)=24x(x^2-2)^3$

Explanation:

$u = x^{2} - 2$ $u=x^2-2$

$z = 3 {(u)}^{4}$ $z=3(u)^4$

$u = x^{2} - 2$ $u=x^2-2$

$\frac{d u}{d x} = 2 x$ $(du)/dx=2x$

$Z = 3 {(u)}^{4}$ $Z=3(u)^4$

$\frac{d z}{d u} = 12 u^{3}$ $(dz)/(du)=12u^3$

we need $\frac{d z}{d x}$ $(dz)/(dx)$

$\frac{d u}{d x} \times \frac{d z}{d u}$ $(du)/dxxx(dz)/(du)$

$\frac{d z}{d x} = 24 x {(x^{2} - 2)}^{3}$ $(dz)/(dx)=24x(x^2-2)^3$

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