What is the solution for -derivative of y=2sin(x)cos(x) ?

The solution for -derivative of y=2sin(x)cos(x) is y=-sin^2(x)+c_{1}

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Find the Derivative - d/dx y=2sin(x)cos(x)

- \frac{d}{d x} y = 2 sin (x) cos (x)

y = - sin^{2} (x) + c_{1}

Solution steps

- \frac{d y}{d x} = 2 sin (x) cos (x)

Isolate

y^{'} (x) : y^{'} (x) = - 2 sin (x) cos (x)

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int - 2 sin (x) cos (x) d x

\int - 2 sin (x) cos (x) d x = - sin^{2} (x) + c_{1}

y = - sin^{2} (x) + c_{1}

- \frac{d}{d x} y = 2 tan (x)

- \frac{d}{d x} (4 sin (x))

- \frac{d}{d x} (sin (x y))

\frac{d y}{d x} y = sin (x)

- \frac{d}{d x} (4)

What is the solution for -derivative of y=2sin(x)cos(x) ?
The solution for -derivative of y=2sin(x)cos(x) is y=-sin^2(x)+c_{1}