What is the d/(dy)(ln(((3y+1)^4)/(sqrt(y^2+1)))) ?

The d/(dy)(ln(((3y+1)^4)/(sqrt(y^2+1)))) is (9y^2-y+12)/((3y+1)(y^2+1))

What is the first d/(dy)(ln(((3y+1)^4)/(sqrt(y^2+1)))) ?

The first d/(dy)(ln(((3y+1)^4)/(sqrt(y^2+1)))) is (9y^2-y+12)/((3y+1)(y^2+1))

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\frac{d}{d y} (ln (\frac{( 3 y + 1 ) ^{4}}{y ^{2} + 1}))

\frac{9 y ^{2} - y + 12}{( 3 y + 1 ) ( y ^{2} + 1 )}

Solution steps

\frac{d}{d y} (ln (\frac{( 3 y + 1 ) ^{4}}{y ^{2} + 1}))

Apply the chain rule:

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

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

\frac{d}{d y} (\frac{( 3 y + 1 ) ^{4}}{y ^{2} + 1}) = \frac{12 ( 3 y + 1 ) ^{3} ( y ^{2} + 1 ) - y ( 3 y + 1 ) ^{4}}{( y ^{2} + 1 ) y ^{2} + 1}

= \frac{1}{\frac{( 3 y + 1 ) ^{4}}{y ^{2} + 1}} \cdot \frac{12 ( 3 y + 1 ) ^{3} ( y ^{2} + 1 ) - y ( 3 y + 1 ) ^{4}}{( y ^{2} + 1 ) y ^{2} + 1}

Simplify

\frac{1}{\frac{( 3 y + 1 ) ^{4}}{y ^{2} + 1}} \cdot \frac{12 ( 3 y + 1 ) ^{3} ( y ^{2} + 1 ) - y ( 3 y + 1 ) ^{4}}{( y ^{2} + 1 ) y ^{2} + 1} : \frac{9 y ^{2} - y + 12}{( 3 y + 1 ) ( y ^{2} + 1 )}

= \frac{9 y ^{2} - y + 12}{( 3 y + 1 ) ( y ^{2} + 1 )}

x \to 0 - lim (x^{2} - \frac{1}{x})

f (x) = \frac{x ^{4}}{4}

\int e^{5 x + 1} d x

y = x, y = x^{2} - 2

\int x^{2} e^{3 x^{3}} d x

What is the d/(dy)(ln(((3y+1)^4)/(sqrt(y^2+1)))) ?
The d/(dy)(ln(((3y+1)^4)/(sqrt(y^2+1)))) is (9y^2-y+12)/((3y+1)(y^2+1))
What is the first d/(dy)(ln(((3y+1)^4)/(sqrt(y^2+1)))) ?
The first d/(dy)(ln(((3y+1)^4)/(sqrt(y^2+1)))) is (9y^2-y+12)/((3y+1)(y^2+1))