What is the derivative of s(t)= 5/(t^2+6t-8) ?

The derivative of s(t)= 5/(t^2+6t-8) is -(5(2t+6))/((t^2+6t-8)^2)

What is the first derivative of s(t)= 5/(t^2+6t-8) ?

The first derivative of s(t)= 5/(t^2+6t-8) is -(5(2t+6))/((t^2+6t-8)^2)

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derivative of

s (t) = \frac{5}{t ^{2} + 6 t - 8}

- \frac{5 ( 2 t + 6 )}{( t ^{2} + 6 t - 8 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{5}{t ^{2} + 6 t - 8})

Take the constant out:

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

= 5 \frac{d}{d t} (\frac{1}{t ^{2} + 6 t - 8})

Apply exponent rule:

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

= 5 \frac{d}{d t} ((t^{2} + 6 t - 8)^{- 1})

Apply the chain rule:

- \frac{1}{( t ^{2} + 6 t - 8 ) ^{2}} \frac{d}{d t} (t^{2} + 6 t - 8)

= - \frac{1}{( t ^{2} + 6 t - 8 ) ^{2}} \frac{d}{d t} (t^{2} + 6 t - 8)

\frac{d}{d t} (t^{2} + 6 t - 8) = 2 t + 6

= 5 (- \frac{1}{( t ^{2} + 6 t - 8 ) ^{2}} (2 t + 6))

Simplify

5 (- \frac{1}{( t ^{2} + 6 t - 8 ) ^{2}} (2 t + 6)) : - \frac{5 ( 2 t + 6 )}{( t ^{2} + 6 t - 8 ) ^{2}}

= - \frac{5 ( 2 t + 6 )}{( t ^{2} + 6 t - 8 ) ^{2}}

\int \frac{x + 4}{x ^{8}} d x

\frac{d}{d x} (7 x e^{- x})

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

x \to 0 lim (\frac{1}{e ^{2 x} - 1})

\int x cos (5 x^{2}) d x

What is the derivative of s(t)= 5/(t^2+6t-8) ?
The derivative of s(t)= 5/(t^2+6t-8) is -(5(2t+6))/((t^2+6t-8)^2)
What is the first derivative of s(t)= 5/(t^2+6t-8) ?
The first derivative of s(t)= 5/(t^2+6t-8) is -(5(2t+6))/((t^2+6t-8)^2)