What is the derivative of 1/(x+y^2) ?

The derivative of 1/(x+y^2) is -(1+2y\frac{dy)/(dx)}{(x+y^2)^2}

What is the first derivative of 1/(x+y^2) ?

The first derivative of 1/(x+y^2) is -(1+2y\frac{dy)/(dx)}{(x+y^2)^2}

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

What is the derivative of 1/(x+y^2) ?
The derivative of 1/(x+y^2) is -(1+2y\frac{dy)/(dx)}{(x+y^2)^2}
What is the first derivative of 1/(x+y^2) ?
The first derivative of 1/(x+y^2) is -(1+2y\frac{dy)/(dx)}{(x+y^2)^2}