What is the (2sin(x)cos(x))^' ?

The (2sin(x)cos(x))^' is cos(2x)*2

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

The first (2sin(x)cos(x))^' is cos(2x)*2

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

cos (2 x) \cdot 2

Solution steps

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

Simplify

2 sin (x) cos (x) : sin (2 x)

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

Apply the chain rule:

cos (2 x) (2 x)^{'}

= cos (2 x) (2 x)^{'}

(2 x)^{'} = 2

= cos (2 x) \cdot 2

\frac{d y}{d x}, 8 cos (6 x) sin (4 y) = 3

\int \frac{25}{9} sin^{2} (x) d x

\int \frac{x ^{2} - 3 x + 7}{x ^{3} + x} d x

f (x) = - 12

\int x^{2} 9 + x^{3} d x