What is the (3^{-x^2})^' ?

The (3^{-x^2})^' is -2ln(3)*3^{-x^2}x

What is the first (3^{-x^2})^' ?

The first (3^{-x^2})^' is -2ln(3)*3^{-x^2}x

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(3^{- x^{2}})^{^{'}}

- 2 ln (3) \cdot 3^{- x^{2}} x

Solution steps

(3^{- x^{2}})^{'}

Apply exponent rule:

a^{b} = e^{b l n (a)}

3^{- x^{2}} = e^{(- x^{2}) l n (3)}

= (e^{(- x^{2}) l n (3)})^{^{'}}

Apply the chain rule:

e^{(- x^{2}) l n (3)} ((- x^{2}) ln (3))^{'}

= e^{(- x^{2}) l n (3)} ((- x^{2}) ln (3))^{'}

((- x^{2}) ln (3))^{'} = - 2 ln (3) x

= e^{(- x^{2}) l n (3)} (- 2 ln (3) x)

Simplify

e^{(- x^{2}) l n (3)} (- 2 ln (3) x) : - 2 ln (3) \cdot 3^{- x^{2}} x

= - 2 ln (3) \cdot 3^{- x^{2}} x

1 - \frac{36}{x ^{2}}

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7 cos (4 x)

- 2 y^{^{''}} + 4 y^{^{'}} + 3 y = 1

f (x) = (x^{2} + 2) (x^{4} + 1)