I was given this following problem.
Question: Let $$y=\sqrt{\cos{x}}.$$ First, show that $$ 4y^2\left(\frac{dy}{dx}\right)^2=1-y^4.$$ And consequently,
$$\frac{d^2y}{dx^2}=-\frac{1}{4}(y^{-3}+y).$$
Find the value of $\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$ at $x=0$.
If I just plow through the calculations (differentiating twice and subbed appropriate terms), I can establish that the expression is indeed correct, and I can use them to calculate the derivatives.
Alternative approach My questions surfaces when I tried another way. If I start from the 2nd equation and manipulated it, I obtain $$\left(\frac{dy}{dx}\right)^2=\frac{1}{4}(y^{-2}-y^2).$$ Differentiating both sides w.r.t $x$ gives $$ 2\frac{dy}{dx}\frac{d^2y}{dx^2}=-\frac{1}{2}(\frac{1}{y^3}+y)\frac{dy}{dx}.$$ Collecting the terms on the LHS gives $$ \frac{dy}{dx} \left(\frac{d^2y}{dx^2}+\frac{1}{4}\left(\frac{1}{y^3}+y\right)\right)=0.$$
At this point, I noticed that I could conclude what I wanted to prove if only $$ \frac{dy}{dx}\neq 0.$$
As $\frac{dy}{dx}=0 $ at $x=0$, I believe that this alternative approach cannot be used to show $\frac{d^2y}{dx^2}=-\frac{1}{4}(y^{-3}+y).$
So, my questions are
- Am I incorrect to conclude that it doesn't work?
- Is there a way to 'fix' this approach?
Thank you in advance for the read and responses.
If $y=\sqrt{g(x)}$, then $y'=\frac{g'(x)}{2\sqrt{g(x)}}$, and since $(\cos x)'=-\sin x$, the first identity is equivalent to: $$ 4\cos x\left(\frac{-\sin x}{2\sqrt{\cos x}}\right)^2 = 1-\cos^2 x$$ that is just $\sin^2 x+\cos^2 x = 1$. Since $y(0)=1$, the relation $$ (2yy')^2 = 1-y^4 \tag{1}$$ gives: $$ (2y'(0))^2 = 1-1=0,$$ hence $y'(0)=0$. Now we have: $$ 8 y' y'' = y'(-2 y^{-3}-2y)$$ and since, due to $(1)$, $y'\neq 0$ in a punctured neighbourhood of the origin $U$, over $U$ we have: $$ 4y'' = -y^{-3}-y\tag{2}$$ so if $y$ is a $C^2$ function in a neighbourhood of the origin, $y''(0)=-\frac{1}{2}$ follows by evaluating $(2)$ in a point $\varepsilon$ near the origin then taking the limit as $\varepsilon\to 0$.