What is expand (1-y)e^yy^'+(y^2)/(xln(x)) ?

The solution to expand (1-y)e^yy^'+(y^2)/(xln(x)) is e^yy^'-e^yy^{1+'}+(y^2)/(xln(x))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand (1-y)e^yy^'+(y^2)/(xln(x))

expand

(1 - y) e^{y} y^{^{'}} + \frac{y ^{2}}{x ln ( x )}

e^{y} y^{'} - e^{y} y^{1 +'} + \frac{y ^{2}}{x ln ( x )}

Solution steps

(1 - y) e^{y} y^{^{'}} + \frac{y ^{2}}{x ln ( x )}

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

Expand

e^{y} y^{^{'}} (1 - y) : e^{y} y^{^{'}} - e^{y} y^{1 +^{'}}

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

(x^{2} - x) (x - 1) 2 x + 1

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

(4 a^{2} ma s^{2} ab, 5 b) ma s (6 a^{2} - 7 ab)

(v + Δ v)^{3}

What is expand (1-y)e^yy^'+(y^2)/(xln(x)) ?
The solution to expand (1-y)e^yy^'+(y^2)/(xln(x)) is e^yy^'-e^yy^{1+'}+(y^2)/(xln(x))