Since we have simplified, we can simply apply the chain rule to it using u = x 3 + 2 x 2 + x − 3 {\displaystyle u=x^{3}+2x^{2}+x-3} .
f ( u ) = e u f ′ ( u ) = e u u ( x ) x 3 + 2 x 2 + x − 3 u ′ ( x ) = 3 x 2 + 4 x + 1. {\displaystyle {\begin{aligned}f(u)&=e^{u}&f'(u)&=e^{u}\\u(x)&x^{3}+2x^{2}+x-3&u'(x)&=3x^{2}+4x+1.\end{aligned}}}
Combining, we get:
f ′ ( x ) = ( 3 x 2 + 4 x + 1 ) ⋅ e x 3 + 2 x 2 + x − 3 {\displaystyle f'(x)=\left(3x^{2}+4x+1\right)\cdot e^{x^{3}+2x^{2}+x-3}}
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