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Popular >

d/(dy)(1/(1+y^2))

Solution

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

Solution

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

Popular Examples

(\partial)/(\partial v)(u+uv)

\frac{\partial}{\partial v} (u + uv)

derivative of y=x^3-8x^2+4x-32 derivative of

y = x^{3} - 8 x^{2} + 4 x - 32

integral of 1/(sin(y))

\int \frac{1}{sin ( y )} d y

integral of sin(x)sec(x)

\int sin (x) sec (x) d x

derivative of sin(x*tan(x))

\frac{d}{d x} (sin (x) \cdot tan (x))

Frequently Asked Questions (FAQ)

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