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

The derivative of y=(5t-1)(2t-3)^{-1} is -(13)/((2t-3)^2)

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

The first derivative of y=(5t-1)(2t-3)^{-1} is -(13)/((2t-3)^2)

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derivative of

y = (5 t - 1) (2 t - 3)^{- 1}

- \frac{13}{( 2 t - 3 ) ^{2}}

Solution steps

\frac{d}{d t} ((5 t - 1) (2 t - 3)^{- 1})

Apply the Product Rule:

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

f = 5 t - 1, g = (2 t - 3)^{- 1}

= \frac{d}{d t} (5 t - 1) (2 t - 3)^{- 1} + \frac{d}{d t} ((2 t - 3)^{- 1}) (5 t - 1)

\frac{d}{d t} (5 t - 1) = 5

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

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

Simplify

5 (2 t - 3)^{- 1} + (- \frac{2}{( 2 t - 3 ) ^{2}}) (5 t - 1) : - \frac{13}{( 2 t - 3 ) ^{2}}

= - \frac{13}{( 2 t - 3 ) ^{2}}

n = 1 \sum \infty \frac{2}{n ( n + 1 )}

\frac{d}{d x} (tan (\frac{5}{x}))

y = 7 x, y = 44 - x^{2}

\int \frac{18}{x ^{2} - 1} d x

n = 2 \sum \infty \frac{1}{ln ( n )}

What is the derivative of y=(5t-1)(2t-3)^{-1} ?
The derivative of y=(5t-1)(2t-3)^{-1} is -(13)/((2t-3)^2)
What is the first derivative of y=(5t-1)(2t-3)^{-1} ?
The first derivative of y=(5t-1)(2t-3)^{-1} is -(13)/((2t-3)^2)