How do you differentiate $f (x) = {(3 x^{3} - 2 x^{2} + 5)}^{331}$ $f(x)=(3x^3-2x^2+5)^331$ ?

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Answer 1 · 2016-10-11T03:40:01

$\frac{d y}{d x} = 331 (9 x^{2} - 4 x) {(3 x^{3} - 2 x^{2} + 5)}^{330}$ $(dy)/(dx)=331(9x^2-4x)(3x^3-2x^2+5)^330$

Using chain rule: $\frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)*(du)/(dx)$

In this case, $y = {(3 x^{3} - 2 x^{2} + 5)}^{331}$ $y=(3x^3-2x^2+5)^331$

Let $u = 3 x^{3} - 2 x^{2} + 5$ $u=3x^3-2x^2+5$ ,

then $\frac{d y}{d u} = 331 u^{330}$ $(dy)/(du)=331u^330$ and $\frac{d u}{d x} = 9 x^{2} - 4 x$ $(du)/(dx)=9x^2-4x$

So $\frac{d y}{d x} = 331 u^{330} \cdot (9 x^{2} - 4 x)$ $(dy)/(dx)=331u^330*(9x^2-4x)$

$= 331 (9 x^{2} - 4 x) {(3 x^{3} - 2 x^{2} + 5)}^{330}$ $=331(9x^2-4x)(3x^3-2x^2+5)^330$

How do you differentiate f(x)=(3x3−2x2+5)331f(x)=(3x^3-2x^2+5)^331?