What is the derivative of f(t)= 1/2 (9t^2+t)^{-3} ?

The derivative of f(t)= 1/2 (9t^2+t)^{-3} is -(3(18t+1))/(2(9t^2+t)^4)

What is the first derivative of f(t)= 1/2 (9t^2+t)^{-3} ?

The first derivative of f(t)= 1/2 (9t^2+t)^{-3} is -(3(18t+1))/(2(9t^2+t)^4)

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derivative of

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

- \frac{3 ( 18 t + 1 )}{2 ( 9 t ^{2} + t ) ^{4}}

Solution steps

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

Take the constant out:

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

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

Apply the chain rule:

- \frac{3}{( 9 t ^{2} + t ) ^{4}} \frac{d}{d t} (9 t^{2} + t)

= - \frac{3}{( 9 t ^{2} + t ) ^{4}} \frac{d}{d t} (9 t^{2} + t)

\frac{d}{d t} (9 t^{2} + t) = 18 t + 1

= \frac{1}{2} (- \frac{3}{( 9 t ^{2} + t ) ^{4}} (18 t + 1))

Simplify

\frac{1}{2} (- \frac{3}{( 9 t ^{2} + t ) ^{4}} (18 t + 1)) : - \frac{3 ( 18 t + 1 )}{2 ( 9 t ^{2} + t ) ^{4}}

= - \frac{3 ( 18 t + 1 )}{2 ( 9 t ^{2} + t ) ^{4}}

x \to \infty lim (2 1^{x})

\frac{\partial}{\partial y} (cos (12 x) sin (2 y))

\int \frac{9}{x ln ( 3 x )} d x

f (x) = - 247

\int \frac{1}{121 + x ^{2}} d x

What is the derivative of f(t)= 1/2 (9t^2+t)^{-3} ?
The derivative of f(t)= 1/2 (9t^2+t)^{-3} is -(3(18t+1))/(2(9t^2+t)^4)
What is the first derivative of f(t)= 1/2 (9t^2+t)^{-3} ?
The first derivative of f(t)= 1/2 (9t^2+t)^{-3} is -(3(18t+1))/(2(9t^2+t)^4)