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

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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a} = a^{- 1}

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

Apply the chain rule:

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

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

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

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

Simplify

x^{2} (- \frac{1}{( y ^{2} + 1 ) ^{2}} \cdot 2 y) : - \frac{2 x ^{2} y}{( y ^{2} + 1 ) ^{2}}

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

f (x) = 4 e^{x} cos (x), at x = 0

y = \frac{5 x ^{2} + 4}{x + 3}

\int_{0}^{\frac{π}{4}} x sin (8 x) d x

x^{2} ln (9 + x)

y = 2 x, (32, 8)

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