What is the d/(dt)((a+bt+ct^2)/(sqrt(t))) ?

The d/(dt)((a+bt+ct^2)/(sqrt(t))) is (3ct^2+bt-a)/(2tsqrt(t))

What is the first d/(dt)((a+bt+ct^2)/(sqrt(t))) ?

The first d/(dt)((a+bt+ct^2)/(sqrt(t))) is (3ct^2+bt-a)/(2tsqrt(t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d t} (\frac{a + b t + c t ^{2}}{t})

\frac{3 c t ^{2} + b t - a}{2 t t}

Solution steps

\frac{d}{d t} (\frac{a + b t + c t ^{2}}{t})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( a + b t + c t ^{2} ) t - \frac{d}{d t} ( t ) ( a + b t + c t ^{2} )}{( t ) ^{2}}

\frac{d}{d t} (a + b t + c t^{2}) = 2 c t + b

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

= \frac{( 2 c t + b ) t - \frac{1}{2 t} ( a + b t + c t ^{2} )}{( t ) ^{2}}

Simplify

\frac{( 2 c t + b ) t - \frac{1}{2 t} ( a + b t + c t ^{2} )}{( t ) ^{2}} : \frac{3 c t ^{2} + b t - a}{2 t t}

= \frac{3 c t ^{2} + b t - a}{2 t t}

\int tan^{6} (3 x) d x

\int 2 3 x - 1 d x

5 \frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 9 x = 2 cos (3 t)

\frac{d}{d x} (7 sin (x) + 7 cos (x))

\int e^{6 x} sin (2 x) d x

What is the d/(dt)((a+bt+ct^2)/(sqrt(t))) ?
The d/(dt)((a+bt+ct^2)/(sqrt(t))) is (3ct^2+bt-a)/(2tsqrt(t))
What is the first d/(dt)((a+bt+ct^2)/(sqrt(t))) ?
The first d/(dt)((a+bt+ct^2)/(sqrt(t))) is (3ct^2+bt-a)/(2tsqrt(t))