What is the d/(dy)((y^4+48)/(24y)) ?

The d/(dy)((y^4+48)/(24y)) is (y^4-16)/(8y^2)

What is the first d/(dy)((y^4+48)/(24y)) ?

The first d/(dy)((y^4+48)/(24y)) is (y^4-16)/(8y^2)

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\frac{d}{d y} (\frac{y ^{4} + 48}{24 y})

\frac{y ^{4} - 16}{8 y ^{2}}

Solution steps

\frac{d}{d y} (\frac{y ^{4} + 48}{24 y})

Take the constant out:

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

= \frac{1}{24} \frac{d}{d y} (\frac{y ^{4} + 48}{y})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{1}{24} \cdot \frac{\frac{d}{d y} ( y ^{4} + 48 ) y - \frac{d y}{d y} ( y ^{4} + 48 )}{y ^{2}}

\frac{d}{d y} (y^{4} + 48) = 4 y^{3}

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

= \frac{1}{24} \cdot \frac{4 y ^{3} y - 1 \cdot ( y ^{4} + 48 )}{y ^{2}}

Simplify

\frac{1}{24} \cdot \frac{4 y ^{3} y - 1 \cdot ( y ^{4} + 48 )}{y ^{2}} : \frac{y ^{4} - 16}{8 y ^{2}}

= \frac{y ^{4} - 16}{8 y ^{2}}

\int \frac{x ^{3} - 1}{x ^{4} - 4 x} d x

\frac{\partial}{\partial x} (\frac{1}{\frac{1}{x} + \frac{1}{y}})

\frac{\partial}{\partial x} (4 arctan (x y))

\int 12 d x

x \to 0 lim (\frac{x}{tan ( 2 x )})

What is the d/(dy)((y^4+48)/(24y)) ?
The d/(dy)((y^4+48)/(24y)) is (y^4-16)/(8y^2)
What is the first d/(dy)((y^4+48)/(24y)) ?
The first d/(dy)((y^4+48)/(24y)) is (y^4-16)/(8y^2)