What is the derivative of (3t-1)(4t-4)^{-1} ?

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

What is the first derivative of (3t-1)(4t-4)^{-1} ?

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

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derivative of

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

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

Solution steps

\frac{d}{d t} ((3 t - 1) (4 t - 4)^{- 1})

Apply the Product Rule:

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

f = 3 t - 1, g = (4 t - 4)^{- 1}

= \frac{d}{d t} (3 t - 1) (4 t - 4)^{- 1} + \frac{d}{d t} ((4 t - 4)^{- 1}) (3 t - 1)

\frac{d}{d t} (3 t - 1) = 3

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

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

Simplify

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

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

\int x \frac{1}{2} d x

\frac{d}{d x} (2 x - 5)

\frac{\partial}{\partial x} (1 + arctan (x))

\int \frac{1}{( x ^{2} - 169 ) ^{\frac{3}{2}}} d x

\int \frac{cos ( x )}{5 + 4 cos ( x )} d x

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