Finding $\frac{d}{dx}\left[ y'(x) \cdot \left(1+y'(x) \right)^{-\frac{1}{2}} \right]$. How do I know what rule to apply?

Question

$\dfrac{d}{dx}\left( \dfrac{y'(x)}{\sqrt{1+y'(x)}} \right)$

I thought of using the normale rules that we have for finding the derivative.

What I obtained is:

$$\dfrac{d}{dx}\left(\dfrac{y'(x)}{\sqrt{1+y'(x)}} \right) = \frac{y''(x)}{\sqrt{1+y'(x)}} - \frac{y'(x)\left(y''(x)\right)^2}{2\sqrt{\left(1+y'(x)^2\right)^3}}$$

Is my working correct? Are there any rules for finding the derivatives in these cases? I too have problems with other similar derivatives, however I do not know which rules to proceed with.

Answer 1

Your first term is correct.

Let's break this down:

$$\left(\frac{a}{b^{\alpha}}\right)'=\frac{a'-\alpha b^{-1}b'a}{b^{\alpha}}=\frac{a'}{b^{\alpha}}-\alpha\frac{b'a}{b^{\alpha+1}}$$ Now $a = y', b=1+y',\alpha=1/2;$ so we have

$$\frac{y''-\frac{1}{2}\frac{y''y'}{1+y'}}{(1+y')^{\frac{1}{2}}}=\frac{y''}{\sqrt{1+y'}}-\frac{1}{2}\frac{y'y''}{\sqrt{(1+y')^3}}.$$