differentiation 4 problem

[red and white kop!]

find d^2/dx^2 as a function of x if siny + cosy = x
i tried to solve this problem this way
differentiating once makes cos y (dy/dx) - sin y (dy/dx) = 1
dy/dx (cos y - siny) = 1
differentiating this makes d^2y/dx^2 (cosy - siny) -dy/dx (sin y + cos y) = 0
d^2y/dx^2 (cos y -siny) - x(dy/dx) = 0
from there i isolate d^2y/dx^2, but apparently its wrong
can anyone help?

 

[Deleted member 4993]

find d^2/dx^2 as a function of x if siny + cosy = x
i tried to solve this problem this way
differentiating once makes cos y (dy/dx) - sin y (dy/dx) = 1
dy/dx (cos y - siny) = 1
differentiating this makes d^2y/dx^2 (cosy - siny) -dy/dx (sin y + cos y) = 0
d^2y/dx^2 (cos y -siny) - x(dy/dx) = 0
from there i isolate d^2y/dx^2, but apparently its wrong
can anyone help?

x = sin(y) + cos(y) = sqrt(2)[sin(?/4 + y)]

sin(?/4 + y) = x/sqrt(2)

y = sin[sup:25u5zbmf]-1[/sup:25u5zbmf][x/sqrt(2)] - ?/4

Now continue....

 

[soroban]

Hello, red and white kop!

Your work is a bit faulty . . . and the final step is elusive.

\(\displaystyle \text{Find }\frac{d^2y}{dx^2}\text{ as a function of }x\text{ if: }\:\sin y + \cos y \:=\: x\)

\(\displaystyle \text{Differentiatie: }\;\cos y \, \frac{dy}{dx} - \sin y\,\frac{dy}{dx} \:=\: 1 \quad\Rightarrow\quad \frac{dy}{dx} (\cos y - \sin y) \:=\: 1\)

. . \(\displaystyle \frac{dy}{dx} \:=\:\frac{1}{\cos y - \sin y} \:=\\cos y - \sin y)^{-1}\)


\(\displaystyle \text{Differentiate again: }\;\frac{d^2y}{dx^2} \:=\:-(\cos y -\sin y)^{-2}(-\sin y - \cos y)\frac{dy}{dx}\)

. . \(\displaystyle \frac{d^2y}{dx^2} \;\;=\;\;\frac{\cos y + \sin y}{(\cos y - \sin y)^2}\cdot\frac{dy}{dx} \;\;=\;\; \frac{x}{(\cos y - \sin y)^2}\cdot\frac{1}{\cos y -\sin y}\)


\(\displaystyle \text{And we have: }\;\frac{d^2y}{dx^2} \;=\;\frac{x}{(\cos y -\sin y)^3} \quad\hdots\;\text{ a function of }x\text{ and }y\)


\(\displaystyle \text{Somehow, we must eliminate the }y\text{ terms . . . but how?}\)


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I found a rather long procedure.
I hope someone can improve on it.


\(\displaystyle \text{We have: }\cos y + \sin y \:=\:x\)

\(\displaystyle \text{Square both sides: }\;\cos^2\!y + 2\sin y\cos y + \sin^2\!y \:=\:x^2\)

. . \(\displaystyle \underbrace{\sin^2\!y + \cos^2\!y}_{\text{This is 1}} + 2\sin y\cos y \:=\:x^2\)

\(\displaystyle \text{We have: }\;1 + 2\sin y\cos y \:=\:x^2 \quad\Rightarrow\quad 2\sin y\cos y \:=\:x^2 - 1\;\;[1]\)


\(\displaystyle \text{Let: }\cos y - \sin y \;=\;k\)

\(\displaystyle \text{Square both sides: }\:\cos^2\!y - 2\sin y\cos y + \sin^2\!y \:=\:k^2 \quad\Rightarrow\quad k^2 \:=\:1 - 2\sin y\cos y\)

\(\displaystyle \text{Substitute [1]: }\;k^2 \:=\:1 - (x^2-1) \:=\:2-x^2 \quad\Rightarrow\quad k \:=\:\sqrt{2-x^2}\)

. . \(\displaystyle \text{Hence: }\;\cos y -\sin y \;=\;\sqrt{2-x^2}\)


\(\displaystyle \text{Therefore: }\;\frac{d^2y}{dx^2} \;=\;\frac{x}{(2-x)^{\frac{3}{2}}}\)

 

[Deleted member 4993]

More direct method

y = sin[sup:296bj987]-1[/sup:296bj987][x/sqrt(2)] - ?/4

\(\displaystyle y = \sin^{-1}\left(\frac{x}{\sqrt{2}}\right ) - \frac{\pi}{4}\)

\(\displaystyle \frac{dy}{dx} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{1-\frac{x^2}{2}}} = \frac{1}{\sqrt{2-x^2}} = (2-x^2)^{-\frac{1}{2}}\)

\(\displaystyle \frac{d^2y}{dx^2} = (-\frac{1}{2})\cdot (-2x)\cdot(2-x^2)^{-\frac{3}{2}} = \frac{x}{(2-x^2)^{\frac{3}{2}}}\)

 

[soroban]

Hello, Subhotosh!

I immediately liked your approach . . . and still do. .It's brilliant!

Yet the problem seemed to demand implicit differentiation.
. . (And very seldom are we able to solve for y.)

And so I felt morally obligated to try it implicitly

 

[BigGlenntheHeavy]

\(\displaystyle cos(y)+sin(y) \ = \ x\)

\(\displaystyle cos^{2}(y)+2sin(y)cos(y)+sin^{2}(y) \ = \ x^{2}\)

\(\displaystyle sin(2y) \ = \ x^{2}-1\)

\(\displaystyle 2y \ = \ arcsin(x^{2}-1)\)

\(\displaystyle y \ = \ \frac{arcsin(x^{2}-1)}{2}\)

\(\displaystyle y' \ = \ \frac{2x}{2\sqrt(1-(x^{2}-1)^{2})} \ = \ \frac{x}{\sqrt(2x^{2}-x^{4})} \ = \ \frac{1}{(2-x^{2})^{1/2}}\)

\(\displaystyle And \ y" \ = \ \frac{x}{(2-x^{2})^{3/2}}\)

\(\displaystyle Note: \ sin(2y) \ = \ x^{2}-1 \ \implies \ -1 \ \le \ x^{2}-1 \ \le \ 1 \ \implies \ |x| \ \le \ \sqrt 2\)

\(\displaystyle Hence, \ we \ have \ three \ critical \ points \ \pm\sqrt 2 \ and \ 0, \ 0 \ snuck \ in \ as \ an \ extraneous \ root.\)

\(\displaystyle Ergo, \ f(\sqrt 2) \ = \ \frac{\pi}{4}, \ f(-\sqrt 2) \ = \ \frac{\pi}{4}, \ and \ f(0) \ = \ -\frac{\pi}{4}, \ no \ inflection\)

\(\displaystyle points, \ see \ graph.\)

[attachment=0:6oshe73a]never.jpg[/attachment:6oshe73a]

 

[red and white kop!]

i dont understand how you differentiated arcsin to get what you got!

 

[BigGlenntheHeavy]

\(\displaystyle \frac{d}{dx}[arcsin(u)] \ = \ \frac{u'}{\sqrt(1-u^{2})}\)

\(\displaystyle Now \ u \ = \ (x^{2}-1), \ u' \ = \ 2x \ and \ u^{2} \ = \ (x^{2}-1)^{2}\)