What is the first (4e^{4x})^' ?

The first (4e^{4x})^' is 16e^{4x}

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(4e^{4x})^'

(4 e^{4 x})^{^{'}}

16 e^{4 x}

Solution steps

(4 e^{4 x})^{'}

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 4 (e^{4 x})^{^{'}}

Apply the chain rule:

e^{4 x} (4 x)^{'}

= e^{4 x} (4 x)^{'}

(4 x)^{'} = 4

= 4 e^{4 x} \cdot 4

Simplify

= 16 e^{4 x}

\frac{x ^{2} + 2 x - 3}{( x + 1 ) ^{2}}

f (x) = \frac{x}{1 + x ^{2}}, at x = 2

\frac{d}{d x} ((x + 2)^{a r c t a n (x)})

\frac{d}{d x} (1 - 2 x - x^{2})

\int_{0}^{1} 162 e^{2 x} + 324 d x