What is the (3x+2sqrt(3x))^' ?

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

What is the first (3x+2sqrt(3x))^' ?

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

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

3 + \frac{3}{x}

Solution steps

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

Simplify

3 x + 2 3 x : 3 x + 23 x

= (3 x + 23 x)^{^{'}}

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= (3 x)^{^{'}} + (23 x)^{^{'}}

(3 x)^{'} = 3

(23 x)^{'} = \frac{3}{x}

= 3 + \frac{3}{x}

x \to \infty lim (0 \cdot x)

\int tan (x) sec^{\frac{3}{2}} (x) d x

\frac{d}{d θ} (cos (4 θ))

\frac{d}{d x} (x (x^{2} - 3))

a re a x^{2} + 3 x - 2, - x^{2} - 3 x + 6