How do you differentiate $r(x)=(e^(41x^2))^3$ ?

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Answer 1 · 2016-10-27T14:45:59

$r'(x) = 246xe^(123x^2)$

We have $r = (e^(41x^2))^3$
$:. r = e^(123x^2)$
$:. (dr)/dx = (123)e^(123x^2)(2x)$
$:. (dr)/dx = 246xe^(123x^2)$

How do you differentiate r(x)=(e^(41x^2))^3r(x)=(e^(41x^2))^3?