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Given ๐‘ข=๐‘ฅ2+๐‘ฅ๐‘ฆ+๐‘ฆ2, solve for โˆ‚๐‘ขโˆ‚๐‘ฅ

Question

Given ๐‘ข=๐‘ฅ2+๐‘ฅ๐‘ฆ+๐‘ฆ2, solve for โˆ‚๐‘ขโˆ‚๐‘ฅ

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Solution

To solve for โˆ‚๐‘ข/โˆ‚๐‘ฅ, we need to take the partial derivative of ๐‘ข with respect to ๐‘ฅ.

The given function is ๐‘ข=๐‘ฅยฒ+๐‘ฅ๐‘ฆ+๐‘ฆยฒ.

The partial derivative of ๐‘ข with respect to ๐‘ฅ is obtained by differentiating each term of ๐‘ข with respect to ๐‘ฅ, while treating all other variables as constants.

  1. The derivative of ๐‘ฅยฒ with respect to ๐‘ฅ is 2๐‘ฅ.
  2. The derivative of ๐‘ฅ๐‘ฆ with respect to ๐‘ฅ is ๐‘ฆ (since the derivative of ๐‘ฅ is 1 and ๐‘ฆ is treated as a constant).
  3. The derivative of ๐‘ฆยฒ with respect to ๐‘ฅ is 0 (since ๐‘ฆยฒ is a constant with respect to ๐‘ฅ).

So, โˆ‚๐‘ข/โˆ‚๐‘ฅ = 2๐‘ฅ + ๐‘ฆ.

This problem has been solved

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