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

The d/(dy)((x(y))((y^4)/4+1/(8y^2))) is (x(10y^6-1))/(8y^2)

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

The first d/(dy)((x(y))((y^4)/4+1/(8y^2))) is (x(10y^6-1))/(8y^2)

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

\frac{x ( 10 y ^{6} - 1 )}{8 y ^{2}}

Solution steps

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

x (1 \cdot (\frac{y ^{4}}{4} + \frac{1}{8 y ^{2}}) + (y^{3} - \frac{1}{4 y ^{3}}) y) : \frac{x ( 10 y ^{6} - 1 )}{8 y ^{2}}

= \frac{x ( 10 y ^{6} - 1 )}{8 y ^{2}}

m^{\frac{1}{3}} (m^{2} - 3 m)

\frac{d ^{2}}{d ^{2} x} (y^{4} + 3 y - 4 x^{3} = 5 x + 1)

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

\frac{( - 9 x + 8 ) ( x - 8 )}{( - 9 x - 8 ) x}

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

What is the d/(dy)((x(y))((y^4)/4+1/(8y^2))) ?
The d/(dy)((x(y))((y^4)/4+1/(8y^2))) is (x(10y^6-1))/(8y^2)
What is the first d/(dy)((x(y))((y^4)/4+1/(8y^2))) ?
The first d/(dy)((x(y))((y^4)/4+1/(8y^2))) is (x(10y^6-1))/(8y^2)