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Suppose I have to find $y’$ in $\sqrt{x+y}=x^3+y^3$. Then instead of differentiating directly, can I first square both sides and then differentiate?

My opinion is that we can’t do it as the original equation is one piece of the equation that we get after squaring. Am I correct on this conclusion? I appreciate if someone can confirm because it is little bit confusing as we can get a similar form for $y’$ for both situations.

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If you square both sides, then you get

$$(\sqrt{x+y})^2=(x^3+y^3)^2$$

Differentiating both sides with respect to $x$:

$$\frac{d(\sqrt{x+y})^2}{dx}=\frac{d(x^3+y^3)^2}{dx}$$

Which by the chain rule can be written as:

$$2\frac{d(\sqrt{x+y})}{dx}=2\frac{d(x^3+y^3)}{dx}$$

which is equivalent to

$$\frac{d(\sqrt{x+y})}{dx}=\frac{d(x^3+y^3)}{dx}.$$

So squaring both sides and then differentiating is fine.

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  • $\begingroup$ I think he wants to get rid of the square root in the left-hand side. $\endgroup$ – wjm Oct 8 '18 at 19:32
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    $\begingroup$ Yes, I know. I am showing it is fine to square both sides to get rid of the square root. $\endgroup$ – smcc Oct 8 '18 at 19:32

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