What is the d/(dy)(y/(t^2+y^2)) ?

The d/(dy)(y/(t^2+y^2)) is (t^2-y^2)/((t^2+y^2)^2)

What is the first d/(dy)(y/(t^2+y^2)) ?

The first d/(dy)(y/(t^2+y^2)) is (t^2-y^2)/((t^2+y^2)^2)

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\frac{d}{d y} (\frac{y}{t ^{2} + y ^{2}})

\frac{t ^{2} - y ^{2}}{( t ^{2} + y ^{2} ) ^{2}}

Solution steps

\frac{d}{d y} (\frac{y}{t ^{2} + y ^{2}})

Treat

t

as a constant

Apply the Quotient Rule:

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

= \frac{\frac{d y}{d y} ( t ^{2} + y ^{2} ) - \frac{d}{d y} ( t ^{2} + y ^{2} ) y}{( t ^{2} + y ^{2} ) ^{2}}

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

\frac{d}{d y} (t^{2} + y^{2}) = 2 y

= \frac{1 \cdot ( t ^{2} + y ^{2} ) - 2 yy}{( t ^{2} + y ^{2} ) ^{2}}

1 \cdot (t^{2} + y^{2}) - 2 yy = t^{2} - y^{2}

= \frac{t ^{2} - y ^{2}}{( t ^{2} + y ^{2} ) ^{2}}

\frac{d r}{d t} + 2 t r - r = 0, r (0) = 6

\frac{\partial}{\partial y} (sin (3 x - 4 y))

\int \frac{1}{x ^{2} + 5} d x

x \to 2 lim (\frac{- 2}{x ( x - 2 )})

f (x) = \frac{e ^{x}}{4 x + 1}

What is the d/(dy)(y/(t^2+y^2)) ?
The d/(dy)(y/(t^2+y^2)) is (t^2-y^2)/((t^2+y^2)^2)
What is the first d/(dy)(y/(t^2+y^2)) ?
The first d/(dy)(y/(t^2+y^2)) is (t^2-y^2)/((t^2+y^2)^2)