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

The first (3^{2x})^' is 2ln(3)*9^x

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(3^{2x})^'

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

2 ln (3) \cdot 9^{x}

Solution steps

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

Simplify

3^{2 x} : 9^{x}

= (9^{x})^{^{'}}

Apply the derivative exponent rule:

\frac{d}{d x} (a^{x}) = a^{x} ln (a)

= 9^{x} ln (9)

Simplify

ln (9) : 2 ln (3)

= 2 ln (3) \cdot 9^{x}

d er i v a t i v e (x - 2)^{4}

\frac{\partial}{\partial x} (e^{- 4 x^{2} - 3 y^{2}})

d er i v a t i v e 5 sec (5 x)

\frac{d}{d x} (ln (5 x + 4))

x \to - 4 lim ((2 x^{3} - 3 x^{2})^{2})