What is the d/(dy)(ln(((2y+1)^5)/(sqrt(y^2+1)))) ?

The d/(dy)(ln(((2y+1)^5)/(sqrt(y^2+1)))) is (8y^2-y+10)/((2y+1)(y^2+1))

What is the first d/(dy)(ln(((2y+1)^5)/(sqrt(y^2+1)))) ?

The first d/(dy)(ln(((2y+1)^5)/(sqrt(y^2+1)))) is (8y^2-y+10)/((2y+1)(y^2+1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

\frac{d}{d y} (ln (\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1}))

\frac{8 y ^{2} - y + 10}{( 2 y + 1 ) ( y ^{2} + 1 )}

Solution steps

\frac{d}{d y} (ln (\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1}))

Apply the chain rule:

\frac{1}{\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1}} \frac{d}{d y} (\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1})

= \frac{1}{\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1}} \frac{d}{d y} (\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1})

\frac{d}{d y} (\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1}) = \frac{10 ( 2 y + 1 ) ^{4} ( y ^{2} + 1 ) - y ( 2 y + 1 ) ^{5}}{( y ^{2} + 1 ) y ^{2} + 1}

= \frac{1}{\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1}} \cdot \frac{10 ( 2 y + 1 ) ^{4} ( y ^{2} + 1 ) - y ( 2 y + 1 ) ^{5}}{( y ^{2} + 1 ) y ^{2} + 1}

Simplify

\frac{1}{\frac{( 2 y + 1 ) ^{5}}{y ^{2} + 1}} \cdot \frac{10 ( 2 y + 1 ) ^{4} ( y ^{2} + 1 ) - y ( 2 y + 1 ) ^{5}}{( y ^{2} + 1 ) y ^{2} + 1} : \frac{8 y ^{2} - y + 10}{( 2 y + 1 ) ( y ^{2} + 1 )}

= \frac{8 y ^{2} - y + 10}{( 2 y + 1 ) ( y ^{2} + 1 )}

What is the d/(dy)(ln(((2y+1)^5)/(sqrt(y^2+1)))) ?
The d/(dy)(ln(((2y+1)^5)/(sqrt(y^2+1)))) is (8y^2-y+10)/((2y+1)(y^2+1))
What is the first d/(dy)(ln(((2y+1)^5)/(sqrt(y^2+1)))) ?
The first d/(dy)(ln(((2y+1)^5)/(sqrt(y^2+1)))) is (8y^2-y+10)/((2y+1)(y^2+1))