What is the d/(dy)(ln(((2y+1)^4)/(sqrt(y^2+1)))) ?

The d/(dy)(ln(((2y+1)^4)/(sqrt(y^2+1)))) is (6y^2-y+8)/((2y+1)(y^2+1))

What is the first d/(dy)(ln(((2y+1)^4)/(sqrt(y^2+1)))) ?

The first d/(dy)(ln(((2y+1)^4)/(sqrt(y^2+1)))) is (6y^2-y+8)/((2y+1)(y^2+1))

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\frac{d}{d y} (ln (\frac{( 2 y + 1 ) ^{4}}{y ^{2} + 1}))

\frac{6 y ^{2} - y + 8}{( 2 y + 1 ) ( y ^{2} + 1 )}

Solution steps

\frac{d}{d y} (ln (\frac{( 2 y + 1 ) ^{4}}{y ^{2} + 1}))

Apply the chain rule:

\frac{1}{\frac{( 2 y + 1 ) ^{4}}{y ^{2} + 1}} \frac{d}{d y} (\frac{( 2 y + 1 ) ^{4}}{y ^{2} + 1})

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

\frac{d}{d y} (\frac{( 2 y + 1 ) ^{4}}{y ^{2} + 1}) = \frac{8 ( 2 y + 1 ) ^{3} ( y ^{2} + 1 ) - y ( 2 y + 1 ) ^{4}}{( y ^{2} + 1 ) y ^{2} + 1}

= \frac{1}{\frac{( 2 y + 1 ) ^{4}}{y ^{2} + 1}} \cdot \frac{8 ( 2 y + 1 ) ^{3} ( y ^{2} + 1 ) - y ( 2 y + 1 ) ^{4}}{( y ^{2} + 1 ) y ^{2} + 1}

Simplify

\frac{1}{\frac{( 2 y + 1 ) ^{4}}{y ^{2} + 1}} \cdot \frac{8 ( 2 y + 1 ) ^{3} ( y ^{2} + 1 ) - y ( 2 y + 1 ) ^{4}}{( y ^{2} + 1 ) y ^{2} + 1} : \frac{6 y ^{2} - y + 8}{( 2 y + 1 ) ( y ^{2} + 1 )}

= \frac{6 y ^{2} - y + 8}{( 2 y + 1 ) ( y ^{2} + 1 )}

\frac{2}{x - 5}

\int \frac{- 1400 x}{25 - x ^{2}} d x

\frac{d}{d x} (sin (x) cos (x) + sin (x))

x^{3} + x^{2}

x \to \infty lim (\frac{m}{n})

What is the d/(dy)(ln(((2y+1)^4)/(sqrt(y^2+1)))) ?
The d/(dy)(ln(((2y+1)^4)/(sqrt(y^2+1)))) is (6y^2-y+8)/((2y+1)(y^2+1))
What is the first d/(dy)(ln(((2y+1)^4)/(sqrt(y^2+1)))) ?
The first d/(dy)(ln(((2y+1)^4)/(sqrt(y^2+1)))) is (6y^2-y+8)/((2y+1)(y^2+1))