What is the first (sin(3x))^' ?

The first (sin(3x))^' is cos(3x)*3

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(sin(3x))^'

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

cos (3 x) \cdot 3

Solution steps

(sin (3 x))^{'}

Apply the chain rule:

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

= cos (3 x) (3 x)^{'}

(3 x)^{'} = 3

= cos (3 x) \cdot 3

x \to 1 lim (4 x - 3)

\int sin (4 x y) d x

(1 - x)^{2}

\frac{d}{d x} (x^{4} (2 x - 5)^{6})

f (x) = \frac{4 x ^{2} - 1}{x ^{3}}