Find $\frac{d^2y}{dx^2}$ as a function of $x$ if $\sin y+\cos y=x$
Ok bit confused as my textbook gives the answer to this problem as:
$$\frac{d^2y}{dx^2}=\pm\frac{x}{\sqrt{(2-x^2)^3}}$$
So I just started in this topic so my methods are kinda basic but what I've done so far is differentiate $\sin y+\cos y=x$ to get:
$$\frac{dy}{dx} = \frac{1}{\cos y-\sin y}$$
But I'm not too sure on how to get the second derivative as $\pm\frac{x}{\sqrt{(2-x^2)^3}}$
Any guidance on this one would be much appreciated.
$\endgroup$4 Answers
$\begingroup$Hint: We have, $$(\sin y + \cos y )^2 = x^2$$
$$\sin 2y = x^2 -1 $$
$$ y = \frac{\arcsin (x^2 -1)}{2} $$
$\endgroup$ $\begingroup$Differentiating implicitly w.r.t $x$ one time yields $$y' = \frac{1}{\cos y - \sin y}$$
Differentiating this implicitly one more time yields $$y'' = \frac{y'(\sin y + \cos y)}{(\cos y - \sin y)^2} = \frac{\sin y + \cos y}{(\cos y- \sin y)^3} = \cdots$$
Combine the above with the fact that $x^2 = 1+2\sin y\cos y \implies 2-x^2 = (\cos y - \sin y)^2$.
$\endgroup$ $\begingroup$You have $$\frac{d^2y}{dx^2}=-(\cos y-\sin y)^{-2}(-\sin y-\cos y)\frac{dy}{dx}$$
$$=\frac{x}{(\cos y-\sin y)^3}$$
Also $$x^2=1+2\sin y\cos y$$ and $$2-x^2=1-2\sin y\cos y=(\cos y-\sin y)^2$$
$\endgroup$ 1 $\begingroup$set $u=y+\frac{\pi}4$ so $u'=y'$ and $u''=y''$ then, since $\sin(y+\frac{\pi}4)= \frac1{\sqrt{2}}(\sin y + \cos y)$ $$ \sin u = \frac1{\sqrt{2}} x \\ u' \cos u = \frac1{\sqrt{2}} \\ u'' \cos u - (u')^2 \sin u =0 $$ hence: $$ u'' = \frac{\sin u}{2^{\frac32}\cos^3 u} \\ = \pm \frac{x}{2^{\frac32}(1-\frac{x^2}2)^\frac32} \\ = \pm\frac{x}{(2-x^2)^{\frac32}} $$
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