What is the derivative of (s^2-8s+15)/(s^2+3s+2) ?

The derivative of (s^2-8s+15)/(s^2+3s+2) is (11s^2-26s-61)/((s^2+3s+2)^2)

What is the first derivative of (s^2-8s+15)/(s^2+3s+2) ?

The first derivative of (s^2-8s+15)/(s^2+3s+2) is (11s^2-26s-61)/((s^2+3s+2)^2)

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derivative of

\frac{s ^{2} - 8 s + 15}{s ^{2} + 3 s + 2}

\frac{11 s ^{2} - 26 s - 61}{( s ^{2} + 3 s + 2 ) ^{2}}

Solution steps

\frac{d}{d s} (\frac{s ^{2} - 8 s + 15}{s ^{2} + 3 s + 2})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d s} ( s ^{2} - 8 s + 15 ) ( s ^{2} + 3 s + 2 ) - \frac{d}{d s} ( s ^{2} + 3 s + 2 ) ( s ^{2} - 8 s + 15 )}{( s ^{2} + 3 s + 2 ) ^{2}}

\frac{d}{d s} (s^{2} - 8 s + 15) = 2 s - 8

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

= \frac{( 2 s - 8 ) ( s ^{2} + 3 s + 2 ) - ( 2 s + 3 ) ( s ^{2} - 8 s + 15 )}{( s ^{2} + 3 s + 2 ) ^{2}}

Expand

(2 s - 8) (s^{2} + 3 s + 2) - (2 s + 3) (s^{2} - 8 s + 15) : 11 s^{2} - 26 s - 61

= \frac{11 s ^{2} - 26 s - 61}{( s ^{2} + 3 s + 2 ) ^{2}}

\frac{d}{d x} (\frac{x ^{2}}{4} + y^{2} - 1)

x y^{^{'}} - 2 y = 11 x^{2}

\frac{d y}{d x} = \frac{- y}{x + y}

\frac{d}{d x} (- \frac{3}{2} x)

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

What is the derivative of (s^2-8s+15)/(s^2+3s+2) ?
The derivative of (s^2-8s+15)/(s^2+3s+2) is (11s^2-26s-61)/((s^2+3s+2)^2)
What is the first derivative of (s^2-8s+15)/(s^2+3s+2) ?
The first derivative of (s^2-8s+15)/(s^2+3s+2) is (11s^2-26s-61)/((s^2+3s+2)^2)