What is the (sqrt(3x+4))^' ?

The (sqrt(3x+4))^' is 3/(2sqrt(3x+4))

What is the first (sqrt(3x+4))^' ?

The first (sqrt(3x+4))^' is 3/(2sqrt(3x+4))

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(3 x + 4)^{^{'}}

\frac{3}{2 3 x + 4}

Solution steps

(3 x + 4)^{'}

Apply the chain rule:

\frac{1}{2 3 x + 4} (3 x + 4)^{'}

= \frac{1}{2 3 x + 4} (3 x + 4)^{'}

(3 x + 4)^{'} = 3

= \frac{1}{2 3 x + 4} \cdot 3

Simplify

\frac{1}{2 3 x + 4} \cdot 3 : \frac{3}{2 3 x + 4}

= \frac{3}{2 3 x + 4}

\int \frac{1}{x ^{4}} \cdot x^{5} e^{x} d x

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

\frac{d}{d x} (e^{x^{2} + 6})

x \to 1 lim (+ (x))

x \to - 6 lim (\frac{x + 7}{x + 6})