What is the derivative of s(t)= 9/(t^2+9t-3) ?

The derivative of s(t)= 9/(t^2+9t-3) is -(9(2t+9))/((t^2+9t-3)^2)

What is the first derivative of s(t)= 9/(t^2+9t-3) ?

The first derivative of s(t)= 9/(t^2+9t-3) is -(9(2t+9))/((t^2+9t-3)^2)

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derivative of

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

- \frac{9 ( 2 t + 9 )}{( t ^{2} + 9 t - 3 ) ^{2}}

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

9 (- \frac{1}{( t ^{2} + 9 t - 3 ) ^{2}} (2 t + 9)) : - \frac{9 ( 2 t + 9 )}{( t ^{2} + 9 t - 3 ) ^{2}}

= - \frac{9 ( 2 t + 9 )}{( t ^{2} + 9 t - 3 ) ^{2}}

\int (\frac{1}{2 x ^{2}}) d x

y^{^{''}} + 3 y^{^{'}} + y = 1

\frac{d}{d x} ((x^{2} + 2 x - 1) \cdot (2 x^{2} - 3 x + 6))

\int 2 e^{- x} d x

h \to 0 lim (\frac{1}{h + 1})

What is the derivative of s(t)= 9/(t^2+9t-3) ?
The derivative of s(t)= 9/(t^2+9t-3) is -(9(2t+9))/((t^2+9t-3)^2)
What is the first derivative of s(t)= 9/(t^2+9t-3) ?
The first derivative of s(t)= 9/(t^2+9t-3) is -(9(2t+9))/((t^2+9t-3)^2)