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

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

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

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

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

cos (2 x)

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = sin (x), g = cos (x)

= (sin (x))^{^{'}} cos (x) + (cos (x))^{^{'}} sin (x)

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

(cos (x))^{'} = - sin (x)

= cos (x) cos (x) + (- sin (x)) sin (x)

Simplify

cos (x) cos (x) + (- sin (x)) sin (x) : cos (2 x)

= cos (2 x)

\int_{1}^{2} 5 x d x

f (x) = 2^{x} 1

\frac{d y}{d t} - 2 t y = 3 t^{2} e^{t^{2}}

\int_{0}^{π} cos^{2} (t) d t