What is the derivative of (sqrt(t))/(1+t^2) ?

The derivative of (sqrt(t))/(1+t^2) is (1-3t^2)/(2sqrt(t)(1+t^2)^2)

What is the first derivative of (sqrt(t))/(1+t^2) ?

The first derivative of (sqrt(t))/(1+t^2) is (1-3t^2)/(2sqrt(t)(1+t^2)^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}{1 + t ^{2}}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int \frac{x ^{4}}{4 x ^{2} + 1} d x

\int_{1}^{2} (x^{2} + 3 x^{4}) d x

x^{2} + 1, x, [- 1, 2]

\frac{d ^{2}}{d x ^{2}} (3 x e^{x})

\int \frac{x - 3}{( x - 5 ) x ^{2}} d x

What is the derivative of (sqrt(t))/(1+t^2) ?
The derivative of (sqrt(t))/(1+t^2) is (1-3t^2)/(2sqrt(t)(1+t^2)^2)
What is the first derivative of (sqrt(t))/(1+t^2) ?
The first derivative of (sqrt(t))/(1+t^2) is (1-3t^2)/(2sqrt(t)(1+t^2)^2)