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shepuryov [24]
2 years ago
11

What the second answer ​

Mathematics
1 answer:
Debora [2.8K]2 years ago
6 0

Answer:

1. y' = 3x² / 4y²

2. y'' = 3x/8y⁵[(4y³ – 3x³)]

Step-by-step explanation:

From the question given above, the following data were obtained:

3x³ – 4y³ = 4

y' =?

y'' =?

1. Determination of y'

To obtain y', we simply defferentiate the expression ones. This can be obtained as follow:

3x³ – 4y³ = 4

Differentiate

9x² – 12y²dy/dx = 0

Rearrange

12y²dy/dx = 9x²

Divide both side by 12y²

dy/dx = 9x² / 12y²

dy/dx = 3x² / 4y²

y' = 3x² / 4y²

2. Determination of y''

To obtain y'', we simply defferentiate above expression i.e y' = 3x² / 4y². This can be obtained as follow:

3x² / 4y²

Let:

u = 3x²

v = 4y²

Find u' and v'

u' = 6x

v' = 8ydy/dx

Applying quotient rule

y'' = [vu' – uv'] / v²

y'' = [4y²(6x) – 3x²(8ydy/dx)] / (4y²)²

y'' = [24xy² – 24x²ydy/dx] / 16y⁴

Recall:

dy/dx = 3x² / 4y²

y'' = [24xy² – 24x²y (3x² / 4y² )] / 16y⁴

y'' = [24xy² – 18x⁴/y] / 16y⁴

y'' = 1/16y⁴[24xy² – 18x⁴/y]

y'' = 1/16y⁴[(24xy³ – 18x⁴)/y]

y'' = 1/16y⁵[(24xy³ – 18x⁴)]

y'' = 6x/16y⁵[(4y³ – 3x³)]

y'' = 3x/8y⁵[(4y³ – 3x³)]

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However, I'm able to rearrange the question as:

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\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

Step-by-step explanation:

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This means that the new second row (R2) is derived by:

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This means that the new third row (R3) is derived by:

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The row 1 elements are:

\left[\begin{array}{ccc}1&1&1|-1\end{array}\right]

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-3 * \left[\begin{array}{ccc}1&1&1|-1\end{array}\right] = \left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right]

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\left[\begin{array}{ccc}-3&-3&-3|3\end{array}\right] + \left[\begin{array}{ccc}3&2&4|1\end{array}\right]

\left[\begin{array}{ccc}0&-1&1|4\end{array}\right]

Hence, the new matrix is:

\left[\begin{array}{ccc}1&1&1|-1\\0&5&7|1\\0&-1&1|4\end{array}\right]

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