What is the derivative of y=(2t-1)(3t-4)^{-1} ?

The derivative of y=(2t-1)(3t-4)^{-1} is -5/((3t-4)^2)

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

The first derivative of y=(2t-1)(3t-4)^{-1} is -5/((3t-4)^2)

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derivative of

y = (2 t - 1) (3 t - 4)^{- 1}

- \frac{5}{( 3 t - 4 ) ^{2}}

Solution steps

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

Apply the Product Rule:

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

f = 2 t - 1, g = (3 t - 4)^{- 1}

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

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

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

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

Simplify

2 (3 t - 4)^{- 1} + (- \frac{3}{( 3 t - 4 ) ^{2}}) (2 t - 1) : - \frac{5}{( 3 t - 4 ) ^{2}}

= - \frac{5}{( 3 t - 4 ) ^{2}}

\int \frac{1 + cos ( 2 θ )}{2} d θ

\frac{\partial}{\partial s} (4 s t)

\frac{d y}{d x} = y - 2 x

x \to 4 lim (x - 6)

x \cdot \frac{d y}{d x} + 3 y = x^{3} - x

What is the derivative of y=(2t-1)(3t-4)^{-1} ?
The derivative of y=(2t-1)(3t-4)^{-1} is -5/((3t-4)^2)
What is the first derivative of y=(2t-1)(3t-4)^{-1} ?
The first derivative of y=(2t-1)(3t-4)^{-1} is -5/((3t-4)^2)