What is the first d/(dt)(t+1/t) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

d/(dt)(t+1/t)

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

1 - \frac{1}{t ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

= 1 - \frac{1}{t ^{2}}

\int (2 a^{2} x^{2} - b^{3})^{2} d x

\int 6 e^{2 t} d t

\int \frac{1}{2} sec^{2} (x) d x

\frac{\partial}{\partial x} (x^{2} y^{3})

\int \frac{4 x}{x ^{3} + x ^{2} - x - 1} d x