What is the (1/(t-1))^' ?

The (1/(t-1))^' is -1/((t-1)^2)

What is the first (1/(t-1))^' ?

The first (1/(t-1))^' is -1/((t-1)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\frac{1}{t - 1})^{^{'}}

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

Solution steps

(\frac{1}{t - 1})^{'}

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= ((t - 1)^{- 1})^{^{'}}

Apply the chain rule:

- \frac{1}{( t - 1 ) ^{2}} (t - 1)^{'}

= - \frac{1}{( t - 1 ) ^{2}} (t - 1)^{'}

(t - 1)^{'} = 1

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

Simplify

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

\int \frac{x}{( 7 x ^{2} + 3 ) ^{5}} d x

y^{^{'''}} + 49 y^{^{'}} = 0

e^{x^{5}}

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

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