What is the derivative of 1/2 (6t^2+t)^{-3} ?

The derivative of 1/2 (6t^2+t)^{-3} is -(3(12t+1))/(2(6t^2+t)^4)

What is the first derivative of 1/2 (6t^2+t)^{-3} ?

The first derivative of 1/2 (6t^2+t)^{-3} is -(3(12t+1))/(2(6t^2+t)^4)

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derivative of

\frac{1}{2} (6 t^{2} + t)^{- 3}

- \frac{3 ( 12 t + 1 )}{2 ( 6 t ^{2} + t ) ^{4}}

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \frac{d}{d t} ((6 t^{2} + t)^{- 3})

Apply the chain rule:

- \frac{3}{( 6 t ^{2} + t ) ^{4}} \frac{d}{d t} (6 t^{2} + t)

= - \frac{3}{( 6 t ^{2} + t ) ^{4}} \frac{d}{d t} (6 t^{2} + t)

\frac{d}{d t} (6 t^{2} + t) = 12 t + 1

= \frac{1}{2} (- \frac{3}{( 6 t ^{2} + t ) ^{4}} (12 t + 1))

Simplify

\frac{1}{2} (- \frac{3}{( 6 t ^{2} + t ) ^{4}} (12 t + 1)) : - \frac{3 ( 12 t + 1 )}{2 ( 6 t ^{2} + t ) ^{4}}

= - \frac{3 ( 12 t + 1 )}{2 ( 6 t ^{2} + t ) ^{4}}

\frac{d}{d x} (e^{3 s i n^{2} (2 x)})

\frac{\partial}{\partial x} (7 e^{x} ln (y))

\frac{d}{d x} (sec^{3} (5 x))

arctan (6 x)

f (x) = x^{3} + 2 x^{2} + 4

What is the derivative of 1/2 (6t^2+t)^{-3} ?
The derivative of 1/2 (6t^2+t)^{-3} is -(3(12t+1))/(2(6t^2+t)^4)
What is the first derivative of 1/2 (6t^2+t)^{-3} ?
The first derivative of 1/2 (6t^2+t)^{-3} is -(3(12t+1))/(2(6t^2+t)^4)