What is the derivative of (s+3)/(4s^2+5) ?

The derivative of (s+3)/(4s^2+5) is (-4s^2-24s+5)/((4s^2+5)^2)

What is the first derivative of (s+3)/(4s^2+5) ?

The first derivative of (s+3)/(4s^2+5) is (-4s^2-24s+5)/((4s^2+5)^2)

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derivative of

\frac{s + 3}{4 s ^{2} + 5}

\frac{- 4 s ^{2} - 24 s + 5}{( 4 s ^{2} + 5 ) ^{2}}

Solution steps

\frac{d}{d s} (\frac{s + 3}{4 s ^{2} + 5})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{\frac{d}{d s} ( s + 3 ) ( 4 s ^{2} + 5 ) - \frac{d}{d s} ( 4 s ^{2} + 5 ) ( s + 3 )}{( 4 s ^{2} + 5 ) ^{2}}

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

\frac{d}{d s} (4 s^{2} + 5) = 8 s

= \frac{1 \cdot ( 4 s ^{2} + 5 ) - 8 s ( s + 3 )}{( 4 s ^{2} + 5 ) ^{2}}

Simplify

\frac{1 \cdot ( 4 s ^{2} + 5 ) - 8 s ( s + 3 )}{( 4 s ^{2} + 5 ) ^{2}} : \frac{- 4 s ^{2} - 24 s + 5}{( 4 s ^{2} + 5 ) ^{2}}

= \frac{- 4 s ^{2} - 24 s + 5}{( 4 s ^{2} + 5 ) ^{2}}

x \to 0 + lim (\frac{1}{x ^{13}})

\frac{\partial}{\partial y} (x - \frac{1}{y})

(- 1.2) (4.3)

y^{^{'}} = - 2 t y^{2}

\int_{0}^{\infty} \frac{1}{3 + x} d x

What is the derivative of (s+3)/(4s^2+5) ?
The derivative of (s+3)/(4s^2+5) is (-4s^2-24s+5)/((4s^2+5)^2)
What is the first derivative of (s+3)/(4s^2+5) ?
The first derivative of (s+3)/(4s^2+5) is (-4s^2-24s+5)/((4s^2+5)^2)