What is the d/(dy)((x(y))((y^4)/8+1/(4y^2))) ?

The d/(dy)((x(y))((y^4)/8+1/(4y^2))) is (x(5y^6-2))/(8y^2)

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

The first d/(dy)((x(y))((y^4)/8+1/(4y^2))) is (x(5y^6-2))/(8y^2)

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Popular >

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

\frac{x ( 5 y ^{6} - 2 )}{8 y ^{2}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y, g = \frac{y ^{4}}{8} + \frac{1}{4 y ^{2}}

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

\frac{d y}{d y} = 1

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

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

Simplify

x (1 \cdot (\frac{y ^{4}}{8} + \frac{1}{4 y ^{2}}) + (\frac{y ^{3}}{2} - \frac{1}{2 y ^{3}}) y) : \frac{x ( 5 y ^{6} - 2 )}{8 y ^{2}}

= \frac{x ( 5 y ^{6} - 2 )}{8 y ^{2}}

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

\frac{( x + 2 ) ^{2}}{4}

(- 1, 3) P (4, - 2)

(\frac{3}{2} - x) (\frac{11}{2} - x)

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

What is the d/(dy)((x(y))((y^4)/8+1/(4y^2))) ?
The d/(dy)((x(y))((y^4)/8+1/(4y^2))) is (x(5y^6-2))/(8y^2)
What is the first d/(dy)((x(y))((y^4)/8+1/(4y^2))) ?
The first d/(dy)((x(y))((y^4)/8+1/(4y^2))) is (x(5y^6-2))/(8y^2)