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

The (sin(x/2))^' is cos(x/2) 1/2

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

The first (sin(x/2))^' is cos(x/2) 1/2

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(sin (\frac{x}{2}))^{^{'}}

cos (\frac{x}{2}) \frac{1}{2}

Solution steps

(sin (\frac{x}{2}))^{'}

Apply the chain rule:

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

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

(\frac{x}{2})^{'} = \frac{1}{2}

= cos (\frac{x}{2}) \frac{1}{2}

\frac{\partial}{\partial x} (x y^{2} + y e^{x^{2}} + 5)

\frac{d y}{d x} + 2 x y = 2 x^{3}, y (0) = 2

\int 1 d ln (x) d x

\frac{d y}{d x} + y = x y^{2}

y = ln (t^{2}) - arctan (\frac{t}{3})