What is the derivative of (6t-1)(4t-4)^{-1} ?

The derivative of (6t-1)(4t-4)^{-1} is -5/(4(t-1)^2)

What is the first derivative of (6t-1)(4t-4)^{-1} ?

The first derivative of (6t-1)(4t-4)^{-1} is -5/(4(t-1)^2)

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derivative of

(6 t - 1) (4 t - 4)^{- 1}

- \frac{5}{4 ( t - 1 ) ^{2}}

Solution steps

\frac{d}{d t} ((6 t - 1) (4 t - 4)^{- 1})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 6 t - 1, g = (4 t - 4)^{- 1}

= \frac{d}{d t} (6 t - 1) (4 t - 4)^{- 1} + \frac{d}{d t} ((4 t - 4)^{- 1}) (6 t - 1)

\frac{d}{d t} (6 t - 1) = 6

\frac{d}{d t} ((4 t - 4)^{- 1}) = - \frac{1}{4 ( t - 1 ) ^{2}}

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

Simplify

6 (4 t - 4)^{- 1} + (- \frac{1}{4 ( t - 1 ) ^{2}}) (6 t - 1) : - \frac{5}{4 ( t - 1 ) ^{2}}

= - \frac{5}{4 ( t - 1 ) ^{2}}

\frac{d}{d x} (x^{4} - 2 a x^{2} - y + 3)

\int 2 x (x^{2} + 10)^{3} d x

\int e^{8 x} cos (5 x) d x

t an g e n t f (x) = 2 e^{x} + x, at x = 0

d er i v a t i v e f (x) = t sin (t)

What is the derivative of (6t-1)(4t-4)^{-1} ?
The derivative of (6t-1)(4t-4)^{-1} is -5/(4(t-1)^2)
What is the first derivative of (6t-1)(4t-4)^{-1} ?
The first derivative of (6t-1)(4t-4)^{-1} is -5/(4(t-1)^2)