What is the first d/(dy)(1-3/y+x) ?

The first d/(dy)(1-3/y+x) is 3/(y^2)

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d/(dy)(1-3/y+x)

\frac{d}{d y} (1 - \frac{3}{y} + x)

\frac{3}{y ^{2}}

Solution steps

\frac{d}{d y} (1 - \frac{3}{y} + x)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (1) - \frac{d}{d y} (\frac{3}{y}) + \frac{d x}{d y}

\frac{d}{d y} (1) = 0

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

\frac{d x}{d y} = 0

= 0 - (- \frac{3}{y ^{2}}) + 0

Simplify

= \frac{3}{y ^{2}}

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

\int \frac{1}{x ( x ^{2} - 49 ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (x \cdot ln (x) - 1)

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

(y^{2} + y x) d x - x^{2} d y = 0