What is the first d/(dy)(2x^3-x^2y) ?

The first d/(dy)(2x^3-x^2y) is -x^2

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d/(dy)(2x^3-x^2y)

\frac{d}{d y} (2 x^{3} - x^{2} y)

- x^{2}

Solution steps

\frac{d}{d y} (2 x^{3} - x^{2} y)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (2 x^{3}) - \frac{d}{d y} (x^{2} y)

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

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

= 0 - x^{2}

Simplify

= - x^{2}

\int \frac{1}{x ^{2} - 12 x + 35} d x

3 x^{^{'}} - 5 x = 12

\frac{d y}{d x} = \frac{1}{2} x + y - 1

y^{^{''''}} + 18 y^{^{''}} + 81 y = 0

x \to 1 lim (x^{2} + 2)