What is the derivative of (sqrt(t))/(2t-3) ?

The derivative of (sqrt(t))/(2t-3) is (-2t-3)/(2sqrt(t)(2t-3)^2)

What is the first derivative of (sqrt(t))/(2t-3) ?

The first derivative of (sqrt(t))/(2t-3) is (-2t-3)/(2sqrt(t)(2t-3)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

\frac{t}{2 t - 3}

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

Solution steps

\frac{d}{d t} (\frac{t}{2 t - 3})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

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

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

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

Simplify

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

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

10 + \frac{20}{x + 1}, at x = 3

x \to \frac{π}{2} lim (cos (π + x))

n = 1 \sum \infty \frac{2}{5}

y^{^{'}} - y^{2} = 0, y (0) = 1

\int_{0}^{1} \frac{x ^{8}}{x ^{9} - 1} d x

What is the derivative of (sqrt(t))/(2t-3) ?
The derivative of (sqrt(t))/(2t-3) is (-2t-3)/(2sqrt(t)(2t-3)^2)
What is the first derivative of (sqrt(t))/(2t-3) ?
The first derivative of (sqrt(t))/(2t-3) is (-2t-3)/(2sqrt(t)(2t-3)^2)