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

Take the constant out:

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

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

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}}

\int - 17 sin^{4} (t) cos (t) d t

\frac{d}{d x} (\frac{x}{3 + x})

\frac{\partial}{\partial t} (cos (t))

x \to 0 lim (4 sin (x) ln (x))

(x^{2} + 1) y^{^{'}} = - 3 x (y - 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)