What is the d/(dt)((3t^2-3)/(2t)) ?

The d/(dt)((3t^2-3)/(2t)) is (3t^2+3)/(2t^2)

What is the first d/(dt)((3t^2-3)/(2t)) ?

The first d/(dt)((3t^2-3)/(2t)) is (3t^2+3)/(2t^2)

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

\frac{3 t ^{2} + 3}{2 t ^{2}}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d}{d t} (3 t^{2} - 3) = 6 t

\frac{d t}{d t} = 1

= \frac{1}{2} \cdot \frac{6 tt - 1 \cdot ( 3 t ^{2} - 3 )}{t ^{2}}

Simplify

\frac{1}{2} \cdot \frac{6 tt - 1 \cdot ( 3 t ^{2} - 3 )}{t ^{2}} : \frac{3 t ^{2} + 3}{2 t ^{2}}

= \frac{3 t ^{2} + 3}{2 t ^{2}}

\frac{d}{d x} (x^{2} + 2 x + 6)

\int_{0}^{1} \frac{27}{4 y - 1} d y

\frac{d ^{2}}{d x ^{2}} (sin (9 x^{2}))

n = 1 \sum \infty 1 0^{n} x^{n} n!

2 x + 5

What is the d/(dt)((3t^2-3)/(2t)) ?
The d/(dt)((3t^2-3)/(2t)) is (3t^2+3)/(2t^2)
What is the first d/(dt)((3t^2-3)/(2t)) ?
The first d/(dt)((3t^2-3)/(2t)) is (3t^2+3)/(2t^2)