What is the d/(dy)(1/t+1/(t^2)-y/(t^2+y^2)) ?

The d/(dy)(1/t+1/(t^2)-y/(t^2+y^2)) is -(t^2-y^2)/((t^2+y^2)^2)

What is the first d/(dy)(1/t+1/(t^2)-y/(t^2+y^2)) ?

The first d/(dy)(1/t+1/(t^2)-y/(t^2+y^2)) is -(t^2-y^2)/((t^2+y^2)^2)

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\frac{d}{d y} (\frac{1}{t} + \frac{1}{t ^{2}} - \frac{y}{t ^{2} + y ^{2}})

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

Solution steps

\frac{d}{d y} (\frac{1}{t} + \frac{1}{t ^{2}} - \frac{y}{t ^{2} + y ^{2}})

Treat

t

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d y} (\frac{1}{t}) = 0

\frac{d}{d y} (\frac{1}{t ^{2}}) = 0

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

= 0 + 0 - \frac{t ^{2} - y ^{2}}{( t ^{2} + y ^{2} ) ^{2}}

Simplify

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

f (x) = 1 - \frac{1}{x ^{2}}

\int x cos (y) d y

x^{2} + y^{2} = 169, (5, 12)

x \to 1 lim (2 - \frac{5}{( x - 1 ) ^{2}})

x (5 x + 3)^{\frac{1}{3}}

What is the d/(dy)(1/t+1/(t^2)-y/(t^2+y^2)) ?
The d/(dy)(1/t+1/(t^2)-y/(t^2+y^2)) is -(t^2-y^2)/((t^2+y^2)^2)
What is the first d/(dy)(1/t+1/(t^2)-y/(t^2+y^2)) ?
The first d/(dy)(1/t+1/(t^2)-y/(t^2+y^2)) is -(t^2-y^2)/((t^2+y^2)^2)