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)

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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}}

\int x (x^{2}) d x

a re a f (x) = 16 - x^{2}, 0, 4

\int \frac{1}{x 64 x ^{2} - 5} d x

\int \frac{3 x ^{3}}{4 + x ^{8}} d x

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3 x} (x + 1)

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)