Respuesta :
Answer:
[tex]\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} - \frac{ {x}^{2} }{ \sqrt{ {(25 - {x}^{2} )}^{3} } } [/tex]
Step-by-step explanation:
[tex]x^2 + y^2 = 25[/tex]
[tex] {y}^{2} = 25 - {x}^{2} [/tex]
[tex]y = \sqrt{(25 - {x}^{2} )} [/tex]
We have to find the double derivative of above equation,
let's find out the first derivative of above equation,
[tex]\frac{dy}{dx} = \frac{d}{dx} \sqrt{(25 - {x}^{2} )} [/tex]
We know that,
[tex] \frac{d}{dx} ( \sqrt{x} ) = \frac{1}{2 \sqrt{x} } [/tex]
but one thing that we should keep in mind, the term inside the root is not x hence we will have to re-diffrentiate the term inside the root.
[tex] \frac{d}{dx} \sqrt{(25 - {x}^{2} )}= \frac{1}{2\sqrt{(25 - {x}^{2} )}} \frac{d}{dx}{(25 - {x}^{2} )}[/tex]
Derivative of any constant number equals zero,
[tex]\frac{d}{dx} \sqrt{(25 - {x}^{2} )} = \frac{1}{2\sqrt{(25 - {x}^{2} )}} ( - 2{x})[/tex]
Simplifying above result,
[tex]\frac{dy}{dx} = \frac{ - {x}}{\sqrt{(25 - {x}^{2} )}} [/tex]
Now let's take the second derivative,
[tex]\frac{ {d}^{2} y}{d {x}^{2} } = \frac{d}{dx} \frac{ - {x}}{\sqrt{(25 - {x}^{2} )}} [/tex]
we can write above term in the form of U.V of derivative
[tex] \frac{d}{dx} U.V = U\frac{d}{dx}V + V\frac{d}{dx}U[/tex]
Similarly,
[tex]\frac{ {d}^{2} y}{d {x}^{2} } = \frac{d}{dx} (x \cdot\frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} )[/tex]
[tex]\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} \frac{d}{dx} x + x\frac{d}{dx}\frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} [/tex]
[tex]\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} + x\frac{d}{dx}\frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} [/tex]
Now we know that,
[tex] \frac{d}{dx} \frac{1}{ \sqrt{x} } = - \frac{1}{2 \sqrt{ {x}^{3} } } [/tex]
[tex]\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} - x \frac{-1}{2 \sqrt{ {(25 - {x}^{2} )}^{3} } } \frac{d}{dx} (25 - {x}^{2} )[/tex]
[tex]\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} - x\frac{-1}{2 \sqrt{ {(25 - {x}^{2} )}^{3} } } ( -2 {x})[/tex]
[tex]\frac{ {d}^{2} y}{d {x}^{2} } = \frac{ - {1}}{\sqrt{(25 - {x}^{2} )}} - \frac{ {x}^{2} }{ \sqrt{ {(25 - {x}^{2} )}^{3} } } [/tex]
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Answer:
[tex]\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = -\dfrac{1}{y} - \dfrac{x^2}{y^3}}[/tex]
Step-by-step explanation:
We are given an equation of circle with radius of 5 units:
[tex]\displaystyle{x^2+y^2=25}[/tex]
To find the second derivative, you'd have to differentiate the equation twice.
We can use implicit differentiation to differentiate. Its concept is to derive like a normal for both sides of equation but since we are differentiating with respect to x and we have y-term, we derive normally then multiply by dy/dx or y'.
For simple clarification, you derive normal and apply chain rules. Hence why there's dy/dx multiplied by 2y:
[tex]\displaystyle{\dfrac{d}{dx}x^2 +\dfrac{d}{dx} y^2 = \dfrac{d}{dx}25}[/tex]
Recall the power rules:
[tex]\displaystyle{\dfrac{d}{dx}ax^n =n\cdot ax^{n-1}}[/tex]
Chain Rules:
[tex]\displaystyle{\dfrac{d}{dx}u^n = n u^{n-1} \cdot \dfrac{du}{dx}}[/tex]
Hence:
[tex]\displaystyle{2x^{2-1} \cdot \dfrac{dx}{dx}+ 2y^{2-1} \cdot \dfrac{dy}{dx} = 0}[/tex]
Note that deriving a constant will always result in 0.
Simplify:
[tex]\displaystyle{2x+ 2y \dfrac{dy}{dx} = 0}[/tex]
Solve for dy/dx:
[tex]\displaystyle{2x+ 2y \dfrac{dy}{dx} = 0}\\\\\displaystyle{2y \dfrac{dy}{dx} = -2x}\\\\\displaystyle{\dfrac{dy}{dx} = \dfrac{-2x}{2y}}\\\\\displaystyle{\dfrac{dy}{dx} = -\dfrac{x}{y}}[/tex]
We've finally found the first derivative of relation. However, we have to find the second derivative, so we derive dy/dx to find the second derivative:
[tex]\displaystyle{\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right) = \dfrac{d^2y}{dx^2}}[/tex]
Therefore:
[tex]\displaystyle{\dfrac{d}{dx}\left(-\dfrac{x}{y}\right)}[/tex]
For this, we will be using quotient rules. Keep in mind that both x and y are function!
Recall quotient rules:
[tex]\displaystyle{\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{u'v - uv'}{v^2}}[/tex]
Let u = -x and v = y:
[tex]\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = \dfrac{(-x)'y - (-x)y'}{y^2}}\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = \dfrac{-1\cdot y - (-x)\cdot \dfrac{dy}{dx}}{y^2}}\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y+x\dfrac{dy}{dx}}{y^2}}[/tex]
Now we know that dy/dx = -x/y, so we substitute dy/dx as -x/y in the second derivative:
[tex]\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y+x\dfrac{dy}{dx}}{y^2}}\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y+x\left(-\dfrac{x}{y}\right)}{y^2}}[/tex]
Simplify:
[tex]\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y-\dfrac{x^2}{y}}{y^2}}[/tex]
More simplification:
[tex]\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right)=\dfrac{-y}{y^2} - \dfrac{\dfrac{x^2}{y}}{y^2}}\\\\\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = -\dfrac{1}{y} - \dfrac{x^2}{y^3}}[/tex]
Therefore, the second derivative is:
[tex]\displaystyle{\dfrac{d}{dx}\left(\dfrac{-x}{y}\right) = -\dfrac{1}{y} - \dfrac{x^2}{y^3}}[/tex]
