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

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

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

y = \frac{1}{x ^{2}}, (3, \frac{1}{9})

\frac{d y}{d x} - \frac{9}{x} y = \frac{y ^{3}}{x ^{13}}

\frac{d}{d x} (\frac{- 7}{x})

f (x) = - x^{2} - x + 4, at x = - 1

y^{^{''}} + 5 y^{^{'}} + 6 y = - 10 t e^{2 t}

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