What is the derivative of 1-3/y+x ?

The derivative of 1-3/y+x is (3\frac{dy)/(dx)}{y^2}+1

What is the first derivative of 1-3/y+x ?

The first derivative of 1-3/y+x is (3\frac{dy)/(dx)}{y^2}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\frac{3 \frac{d y}{d x}}{y ^{2}} + 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

\frac{d x}{d x} = 1

= 0 - (- \frac{3 \frac{d y}{d x}}{y ^{2}}) + 1

Simplify

= \frac{3 \frac{d y}{d x}}{y ^{2}} + 1

What is the derivative of 1-3/y+x ?
The derivative of 1-3/y+x is (3\frac{dy)/(dx)}{y^2}+1
What is the first derivative of 1-3/y+x ?
The first derivative of 1-3/y+x is (3\frac{dy)/(dx)}{y^2}+1