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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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}}