What is the first d/(dy)((x^2)/2-xy) ?

The first d/(dy)((x^2)/2-xy) is -x

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d/(dy)((x^2)/2-xy)

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

- x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

= 0 - x

Simplify

= - x

4 x^{2} - 2 x + 6

\int \frac{2 + 3 sin ^{2} ( x )}{sin ^{2} ( x )} d x

y^{^{'}} + \frac{4}{x ^{2} - 1} y = 0

\int 4 t^{2} d t

x \to 1 lim (- \frac{1}{x - 1})