Question

How do you find `(d^2y)/(dx^2)` for `5y^2+2=5x^2`?

Answer 1

`(d^2y)/dx^2=(y^2-x^2)/y^3`

Explanation:

To find `(d^2y)/(dx^2)` is to find the second differentiation of the
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given expression .
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`5y^2+2=5x^2`
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`d/dx(5y^2+2)=d/dx(5x^2)`
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`rArr10y(dy/dx)+0=10x`
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`rArr10y(dy/dx)=10x`
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`rArrdy/dx=(10x)/(10y)`
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Therefore,
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`color(blue)(dy/dx=x/y`
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Let us compute `(d^2y)/dx^2`
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`(d^2y)/dx^2=d/dx(dy/dx)`
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`(d^2y)/dx^2=d/dx(x/y)`
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Differentiating `x/y` is determined by applying the quotient rule differentiation.
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`(d^2y)/dx^2=((dx/dx)xxy-(dy/dx)xx x)/y^2`
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`(d^2y)/dx^2=(y-xcolor(blue)(dy/dx))/y^2`
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`(d^2y)/dx^2=(y-x(x/y))/y^2`
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`(d^2y)/dx^2=(y-x^2/y)/y^2`
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Therefore,
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`(d^2y)/dx^2=(y^2-x^2)/y^3`

Answer 2

The answer is `=-2/5((5x^2-2)/5)^(-3/2)`

Explanation:

Let's do the implicit differentiation

`5y^2+2=5x^2`

`10ydy/dx+0=10x`

`dy/dx=x/y`

This is of the form `u/v`

`(u/v)'=(u'v-uv')/v^2`

So, `(d^2y)/dx^2=(x/y)'=(y-xdy/dx)/y^2`

`=(y-x^2/y)/y^2=(y^2-x^2)/y^3`

`5y^2=5x^2-2`

`y^2=(5x^2-2)/5`

`y^3=((5x^2-2)/5)^(3/2)`

Therefore, `(d^2y)/dx^2=((5x^2-2)/5-x^2)/((5x^2-2)/5)^(3/2)`

`=-2/(5((5x^2-2)/5)^(3/2))`

`=-2/5((5x^2-2)/5)^(-3/2)`