What is the d/(dt)((2t)/(sqrt(1+4t^2))) ?

The d/(dt)((2t)/(sqrt(1+4t^2))) is 2/((1+4t^2)sqrt(1+4t^2))

What is the first d/(dt)((2t)/(sqrt(1+4t^2))) ?

The first d/(dt)((2t)/(sqrt(1+4t^2))) is 2/((1+4t^2)sqrt(1+4t^2))

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{2 t}{1 + 4 t ^{2}})

\frac{2}{( 1 + 4 t ^{2} ) 1 + 4 t ^{2}}

Solution steps

\frac{d}{d t} (\frac{2 t}{1 + 4 t ^{2}})

Take the constant out:

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

= 2 \frac{d}{d t} (\frac{t}{1 + 4 t ^{2}})

Apply the Quotient Rule:

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

= 2 \cdot \frac{\frac{d t}{d t} 1 + 4 t ^{2} - \frac{d}{d t} ( 1 + 4 t ^{2} ) t}{( 1 + 4 t ^{2} ) ^{2}}

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

\frac{d}{d t} (1 + 4 t^{2}) = \frac{4 t}{1 + 4 t ^{2}}

= 2 \cdot \frac{1 \cdot 1 + 4 t ^{2} - \frac{4 t}{1 + 4 t ^{2}} t}{( 1 + 4 t ^{2} ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot 1 + 4 t ^{2} - \frac{4 t}{1 + 4 t ^{2}} t}{( 1 + 4 t ^{2} ) ^{2}} : \frac{2}{( 1 + 4 t ^{2} ) 1 + 4 t ^{2}}

= \frac{2}{( 1 + 4 t ^{2} ) 1 + 4 t ^{2}}

\int_{0}^{1} π (2 x^{2})^{2} d x

y^{^{''}} + 4 y = 2 sec (2 t)

4 x^{2} + 5 x, at x = 3

\frac{\partial}{\partial x} (e^{x y} + x + y)

\int tan^{5} (x) d x

What is the d/(dt)((2t)/(sqrt(1+4t^2))) ?
The d/(dt)((2t)/(sqrt(1+4t^2))) is 2/((1+4t^2)sqrt(1+4t^2))
What is the first d/(dt)((2t)/(sqrt(1+4t^2))) ?
The first d/(dt)((2t)/(sqrt(1+4t^2))) is 2/((1+4t^2)sqrt(1+4t^2))