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

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

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

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

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

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

Solution steps

(4 + 3 x)^{'}

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\int x^{4} \cdot e^{- x^{5}} d x

x \to 0 lim (\frac{cos ( a x ) - cos ( b x )}{x ^{2}})

f (x) = 5 x^{e}

y^{^{'}} = 1 - sin (y), y (0) = 0