What is implicit (d^2y)/(dx^2),x^2+y^2=16 ?

The answer to implicit (d^2y)/(dx^2),x^2+y^2=16 is (d^2y)/(dx^2)=-(y^2+x^2)/(y^3)

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implicit (d^2y)/(dx^2),x^2+y^2=16

implicit derivative

\frac{d ^{2} y}{d x ^{2}}, x^{2} + y^{2} = 16

\frac{d ^{2} y}{d x ^{2}} = - \frac{y ^{2} + x ^{2}}{y ^{3}}

Solution steps

x^{2} + y^{2} = 16

Treat

y

y (x)

Implicit Derivative

\frac{d y}{d x}

x^{2} + y^{2} = 16 : - \frac{x}{y}

\frac{d y}{d x} = - \frac{x}{y}

Implicit Derivative

\frac{d ^{2} y}{d x ^{2}}

\frac{d y}{d x} = - \frac{x}{y} : - \frac{y - x \frac{d y}{d x}}{y ^{2}}

\frac{d ^{2} y}{d x ^{2}} = - \frac{y - x \frac{d y}{d x}}{y ^{2}}

Substitute

\frac{d y}{d x} = - \frac{x}{y} : - \frac{y ^{2} + x ^{2}}{y ^{3}}

\frac{d ^{2} y}{d x ^{2}} = - \frac{y ^{2} + x ^{2}}{y ^{3}}

\frac{\partial}{\partial z} (x y + yz - x z)

1 - y^{2} d x = 1 - x^{2} d y

\int_{0}^{2} \frac{2 x}{( x ^{2} + 5 ) ^{3}} d x

x \to 0 lim (\frac{- 2 e ^{- x}}{2 x - 2})

\int \frac{e ^{4 x}}{- 4 x} d x

What is implicit (d^2y)/(dx^2),x^2+y^2=16 ?
The answer to implicit (d^2y)/(dx^2),x^2+y^2=16 is (d^2y)/(dx^2)=-(y^2+x^2)/(y^3)