Question
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nd hence, \( (x+)^{\prime}= \) From here let's rearrange algebraically: a) Take the square roots ot both sides, don't forget to indicate there are two answers b) Simplity the fiaction by recalling \( \sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}} \)

Ask by Reid Franklin. in South Africa
Mar 12,2025

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Answer

\( x' = \pm \frac{\sqrt{x}}{\sqrt{y}} \)

Solution

1. Assume we begin with an equation of the form \[ (x')^2 = \frac{x}{y}. \] 2. Taking the square root of both sides (and remembering there are two possible answers), we obtain \[ x' = \pm\sqrt{\frac{x}{y}}. \] 3. Finally, we simplify the right‐side by applying the property \[ \sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}, \] so that \[ x' = \pm \frac{\sqrt{x}}{\sqrt{y}}. \]

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Simplify this solution

Mind Expander

When you differentiate \( (x + k)^{\prime} \), where \( k \) is a constant, the derivative with respect to \( x \) is simply \( 1 \). Differentiating gives you a nice, clean result because the derivative of a constant is zero, leaving only the coefficient of \( x \). Now, moving on to rearranging algebraically! When you take the square root of both sides, remember that square roots can be both positive and negative. So you end up with two solutions: \( x = \pm \sqrt{y} \). And don't forget to simplify those fractions correctly; using the property \( \sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}} \), you can express your final answer clearly and concisely, making math feel like a walk in the park!

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