What is the d/(dy)((x)((y^2+2y)/8-ln(y+1))) ?

The d/(dy)((x)((y^2+2y)/8-ln(y+1))) is (x(y^2+2y-3))/(4(y+1))

What is the first d/(dy)((x)((y^2+2y)/8-ln(y+1))) ?

The first d/(dy)((x)((y^2+2y)/8-ln(y+1))) is (x(y^2+2y-3))/(4(y+1))

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\frac{d}{d y} ((x) (\frac{y ^{2} + 2 y}{8} - ln (y + 1)))

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

Solution steps

\frac{d}{d y} ((x) (\frac{y ^{2} + 2 y}{8} - ln (y + 1)))

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= x \frac{d}{d y} (\frac{y ^{2} + 2 y}{8} - ln (y + 1))

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= x (\frac{d}{d y} (\frac{y ^{2} + 2 y}{8}) - \frac{d}{d y} (ln (y + 1)))

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

\frac{d}{d y} (ln (y + 1)) = \frac{1}{y + 1}

= x (\frac{y + 1}{4} - \frac{1}{y + 1})

Simplify

x (\frac{y + 1}{4} - \frac{1}{y + 1}) : \frac{x ( y ^{2} + 2 y - 3 )}{4 ( y + 1 )}

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

\frac{3 ( 3 3 + 43 )}{4}

\frac{12 h}{14 h}

\frac{2 s ^{2} + 16 s + 34}{( s + 4 ) ( s ^{2} + 4 s + 13 )}

x^{5} (14 (1 + 2 x)^{6})

\frac{ab}{( - b x + a ) ( b x + a )}

What is the d/(dy)((x)((y^2+2y)/8-ln(y+1))) ?
The d/(dy)((x)((y^2+2y)/8-ln(y+1))) is (x(y^2+2y-3))/(4(y+1))
What is the first d/(dy)((x)((y^2+2y)/8-ln(y+1))) ?
The first d/(dy)((x)((y^2+2y)/8-ln(y+1))) is (x(y^2+2y-3))/(4(y+1))