What is the d/(dv)(u/(u+v)) ?

The d/(dv)(u/(u+v)) is -u/((u+v)^2)

What is the first d/(dv)(u/(u+v)) ?

The first d/(dv)(u/(u+v)) is -u/((u+v)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d v} (\frac{u}{u + v})

- \frac{u}{( u + v ) ^{2}}

Solution steps

\frac{d}{d v} (\frac{u}{u + v})

Treat

u

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= u \frac{d}{d v} (\frac{1}{u + v})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= u \frac{d}{d v} ((u + v)^{- 1})

Apply the chain rule:

- \frac{1}{( u + v ) ^{2}} \frac{d}{d v} (u + v)

= - \frac{1}{( u + v ) ^{2}} \frac{d}{d v} (u + v)

\frac{d}{d v} (u + v) = 1

= u (- \frac{1}{( u + v ) ^{2}} \cdot 1)

Simplify

u (- \frac{1}{( u + v ) ^{2}} \cdot 1) : - \frac{u}{( u + v ) ^{2}}

= - \frac{u}{( u + v ) ^{2}}

- x^{3} - 5 x^{2} + 2, at x = - 1

\int cos (x) + 2 d x

\frac{d}{d x} (6 arcsin (x^{2}))

f (x) = x - \frac{1}{x ^{2}}

\int 9 + 9 tan^{2} (x) d x