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

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

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

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

y = cos (x) + c_{1}

Solution steps

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

Isolate

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

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

then

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

y = \int - sin (x) d x

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

y = cos (x) + c_{1}

f (x) = cos (2 x)

- \frac{d}{d} y - 2 y

- \frac{d}{d x} (2 x)

- \frac{d}{d x} (sin (3 x))

- \frac{d}{d x} (2 cos (x))

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