What is the (e^{1/x}-(e^{1/x})/x-(5e^{1/x})/(x^2))^' ?

The (e^{1/x}-(e^{1/x})/x-(5e^{1/x})/(x^2))^' is (11e^{1/x}x+5e^{1/x})/(x^4)

What is the first (e^{1/x}-(e^{1/x})/x-(5e^{1/x})/(x^2))^' ?

The first (e^{1/x}-(e^{1/x})/x-(5e^{1/x})/(x^2))^' is (11e^{1/x}x+5e^{1/x})/(x^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(e^{\frac{1}{x}} - \frac{e ^{\frac{1}{x}}}{x} - \frac{5 e ^{\frac{1}{x}}}{x ^{2}})^{^{'}}

\frac{11 e ^{\frac{1}{x}} x + 5 e ^{\frac{1}{x}}}{x ^{4}}

Solution steps

(e^{\frac{1}{x}} - \frac{e ^{\frac{1}{x}}}{x} - \frac{5 e ^{\frac{1}{x}}}{x ^{2}})^{'}

Apply the Sum/Difference Rule:

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

= (e^{\frac{1}{x}})^{^{'}} - (\frac{e ^{\frac{1}{x}}}{x})^{^{'}} - (\frac{5 e ^{\frac{1}{x}}}{x ^{2}})^{^{'}}

(e^{\frac{1}{x}})^{'} = - \frac{e ^{\frac{1}{x}}}{x ^{2}}

(\frac{e ^{\frac{1}{x}}}{x})^{'} = \frac{- e ^{\frac{1}{x}} - e ^{\frac{1}{x}} x}{x ^{3}}

(\frac{5 e ^{\frac{1}{x}}}{x ^{2}})^{'} = \frac{5 ( - e ^{\frac{1}{x}} - 2 e ^{\frac{1}{x}} x )}{x ^{4}}

= - \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{- e ^{\frac{1}{x}} - e ^{\frac{1}{x}} x}{x ^{3}} - \frac{5 ( - e ^{\frac{1}{x}} - 2 e ^{\frac{1}{x}} x )}{x ^{4}}

Simplify

- \frac{e ^{\frac{1}{x}}}{x ^{2}} - \frac{- e ^{\frac{1}{x}} - e ^{\frac{1}{x}} x}{x ^{3}} - \frac{5 ( - e ^{\frac{1}{x}} - 2 e ^{\frac{1}{x}} x )}{x ^{4}} : \frac{11 e ^{\frac{1}{x}} x + 5 e ^{\frac{1}{x}}}{x ^{4}}

= \frac{11 e ^{\frac{1}{x}} x + 5 e ^{\frac{1}{x}}}{x ^{4}}

t \to 0 lim ((t - 2)^{5})

\frac{( x - 1 )}{( x + 2 )}

y^{^{'}} = \frac{4 x ln ( 2 x )}{sin ( y )}

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

y = \frac{1 - x}{1 + x}

What is the (e^{1/x}-(e^{1/x})/x-(5e^{1/x})/(x^2))^' ?
The (e^{1/x}-(e^{1/x})/x-(5e^{1/x})/(x^2))^' is (11e^{1/x}x+5e^{1/x})/(x^4)
What is the first (e^{1/x}-(e^{1/x})/x-(5e^{1/x})/(x^2))^' ?
The first (e^{1/x}-(e^{1/x})/x-(5e^{1/x})/(x^2))^' is (11e^{1/x}x+5e^{1/x})/(x^4)