What is the (sin(t)cos(t))^' ?

The (sin(t)cos(t))^' is cos(2t)

What is the first (sin(t)cos(t))^' ?

The first (sin(t)cos(t))^' is cos(2t)

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(sin (t) cos (t))^{^{'}}

cos (2 t)

Solution steps

(sin (t) cos (t))^{'}

Apply the Product Rule:

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

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

= (sin (t))^{^{'}} cos (t) + (cos (t))^{^{'}} sin (t)

(sin (t))^{'} = cos (t)

(cos (t))^{'} = - sin (t)

= cos (t) cos (t) + (- sin (t)) sin (t)

Simplify

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

= cos (2 t)

\frac{d ^{5}}{d x ^{5}} (x^{\frac{11}{2}})

x = 8 - y^{2}, x = y^{2}

\int x + 1 d x

x \to 0 lim (\frac{4}{x ^{3} ( x + 3 )})

sin (x), 1