What is the d/(ds)(3/(s^2+9)) ?

The d/(ds)(3/(s^2+9)) is -(6s)/((s^2+9)^2)

What is the first d/(ds)(3/(s^2+9)) ?

The first d/(ds)(3/(s^2+9)) is -(6s)/((s^2+9)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d s} (\frac{3}{s ^{2} + 9})

- \frac{6 s}{( s ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d s} (\frac{3}{s ^{2} + 9})

Take the constant out:

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

= 3 \frac{d}{d s} (\frac{1}{s ^{2} + 9})

Apply exponent rule:

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

= 3 \frac{d}{d s} ((s^{2} + 9)^{- 1})

Apply the chain rule:

- \frac{1}{( s ^{2} + 9 ) ^{2}} \frac{d}{d s} (s^{2} + 9)

= - \frac{1}{( s ^{2} + 9 ) ^{2}} \frac{d}{d s} (s^{2} + 9)

\frac{d}{d s} (s^{2} + 9) = 2 s

= 3 (- \frac{1}{( s ^{2} + 9 ) ^{2}} \cdot 2 s)

Simplify

3 (- \frac{1}{( s ^{2} + 9 ) ^{2}} \cdot 2 s) : - \frac{6 s}{( s ^{2} + 9 ) ^{2}}

= - \frac{6 s}{( s ^{2} + 9 ) ^{2}}

\int arctan (x) d x

cos^{2} (t) \frac{d y}{d t} = 1

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

\frac{\partial}{\partial x} (2 x^{2} + 3 y^{2})

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

What is the d/(ds)(3/(s^2+9)) ?
The d/(ds)(3/(s^2+9)) is -(6s)/((s^2+9)^2)
What is the first d/(ds)(3/(s^2+9)) ?
The first d/(ds)(3/(s^2+9)) is -(6s)/((s^2+9)^2)