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

The first (sin^2(3x))^' is 3sin(6x)

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(sin^2(3x))^'

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

3 sin (6 x)

Solution steps

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

Apply the chain rule:

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

= 2 sin (3 x) (sin (3 x))^{'}

(sin (3 x))^{'} = cos (3 x) \cdot 3

= 2 sin (3 x) cos (3 x) \cdot 3

Simplify

2 sin (3 x) cos (3 x) \cdot 3 : 3 sin (6 x)

= 3 sin (6 x)

\int \frac{5 - 50 x}{9 - 25 x ^{2}} d x

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

f (x) = 2 - 3 x

\int - 7 sec^{2} (x) d x

y^{^{'}} + 5 y = 0, y (0) = 2