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I know dy / dx = n^n-1

I have the problem $y = \sqrt{x^2+2x}$

I have broken that down to $y = \sqrt{x^2 +2x}$. The x would differentiate to 1, but how do you differentiate the surd?

jordan b
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  • please check my formulae editing. If I have guessed it correctly, then you are wrong, $\sqrt{x^2+2x}\neq x + \sqrt{2x}$ – Yulia V Oct 02 '14 at 20:39

2 Answers2

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First note that the chain rule is $$ \frac{d}{dx}f(g(x))= \frac{d}{dg(x)}f(g(x)) \frac{d}{dx}g(x) $$ Therefore $$ \frac{d}{dx}\sqrt{x^2+2x}= \frac{d}{dx}(x^2+2x)^{\frac{1}{2}}= \frac{1}{2}(x^2+2x)^{\frac{1}{2}-1}\frac{d}{dx}[x^2+2x] =\frac{1}{2}(x^2+2x)^{-\frac{1}{2}}(2x+2) = \frac{2x+2}{2(x^2+2x)^{\frac{1}{2}}} = \frac{x+1}{\sqrt{x^2+2x}} $$

k170
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you must write $(x^2+2x)^{1/2}$ and the derivative with respect to $x$ is given by $\frac{1}{2}(x^2+2x)^{-1/2}(2x+2)$

Dr. Sonnhard Graubner
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