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))

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\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 )}

ln (1 + 3 x)

\frac{d}{d x} (\frac{1}{3} x^{3} + \frac{1}{4 x})

y^{^{'}} = 2 + 3 x + 8 y + 12 x y

- 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}

\frac{d}{d x} (e^{- 2 x} y)

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))