How do you find the derivative of $3^(x^2-2)$ ?

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Answer 1 · 2016-06-22T19:54:46

$ln 9 \ x \ 3^(x^2-2)$

let $y = 3^(x^2-2)$

so $ln y = ln (3^(x^2-2)) = (x^2 - 2) ln 3$

$\implies 1/y y' = 2x ln 3$

$y' = 2x ln 3 * y = 2 \ ln 3 \ x \ 3^(x^2-2)$

$= ln 9 \ x \ 3^(x^2-2)$

How do you find the derivative of 3^(x^2-2)3^(x^2-2)?