differentiation 4 problem

R

red and white kop!

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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?
 
red and white kop! said:
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?
Click to expand...

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....
 
Hello, red and white kop!

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



Differentiatie:   cos⁡y dydx−sin⁡y dydx = 1⇒dydx(cos⁡y−sin⁡y) = 1\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

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


Differentiate again:   d2ydx2 = −(cos⁡y−sin⁡y)−2(−sin⁡y−cos⁡y)dydx\displaystyle \text{Differentiate again: }\;\frac{d^2y}{dx^2} \:=\:-(\cos y -\sin y)^{-2}(-\sin y - \cos y)\frac{dy}{dx}

. . d2ydx2    =    cos⁡y+sin⁡y(cos⁡y−sin⁡y)2⋅dydx    =    x(cos⁡y−sin⁡y)2⋅1cos⁡y−sin⁡y\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}


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


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


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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


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

Square both sides:   cos⁡2 ⁣y+2sin⁡ycos⁡y+sin⁡2 ⁣y = x2\displaystyle \text{Square both sides: }\;\cos^2\!y + 2\sin y\cos y + \sin^2\!y \:=\:x^2

. . sin⁡2 ⁣y+cos⁡2 ⁣y⏟This is 1+2sin⁡ycos⁡y = x2\displaystyle \underbrace{\sin^2\!y + \cos^2\!y}_{\text{This is 1}} + 2\sin y\cos y \:=\:x^2

We have:   1+2sin⁡ycos⁡y = x2⇒2sin⁡ycos⁡y = x2−1    [1]\displaystyle \text{We have: }\;1 + 2\sin y\cos y \:=\:x^2 \quad\Rightarrow\quad 2\sin y\cos y \:=\:x^2 - 1\;\;[1]


Let: cos⁡y−sin⁡y  =  k\displaystyle \text{Let: }\cos y - \sin y \;=\;k

Square both sides:  cos⁡2 ⁣y−2sin⁡ycos⁡y+sin⁡2 ⁣y = k2⇒k2 = 1−2sin⁡ycos⁡y\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

Substitute [1]:   k2 = 1−(x2−1) = 2−x2⇒k = 2−x2\displaystyle \text{Substitute [1]: }\;k^2 \:=\:1 - (x^2-1) \:=\:2-x^2 \quad\Rightarrow\quad k \:=\:\sqrt{2-x^2}

. . Hence:   cos⁡y−sin⁡y  =  2−x2\displaystyle \text{Hence: }\;\cos y -\sin y \;=\;\sqrt{2-x^2}


Therefore:   d2ydx2  =  x(2−x)32\displaystyle \text{Therefore: }\;\frac{d^2y}{dx^2} \;=\;\frac{x}{(2-x)^{\frac{3}{2}}}

 
More direct method

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

y=sin⁡−1(x2)−π4\displaystyle y = \sin^{-1}\left(\frac{x}{\sqrt{2}}\right ) - \frac{\pi}{4}

dydx=121−x22=12−x2=(2−x2)−12\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}}

d2ydx2=(−12)⋅(−2x)⋅(2−x2)−32=x(2−x2)32\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}}}
 
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

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

cos2(y)+2sin(y)cos(y)+sin2(y) = x2\displaystyle cos^{2}(y)+2sin(y)cos(y)+sin^{2}(y) \ = \ x^{2}

sin(2y) = x2−1\displaystyle sin(2y) \ = \ x^{2}-1

2y = arcsin(x2−1)\displaystyle 2y \ = \ arcsin(x^{2}-1)

y = arcsin(x2−1)2\displaystyle y \ = \ \frac{arcsin(x^{2}-1)}{2}

y′ = 2x2(1−(x2−1)2) = x(2x2−x4) = 1(2−x2)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}}

And y" = x(2−x2)3/2\displaystyle And \ y" \ = \ \frac{x}{(2-x^{2})^{3/2}}

Note: sin(2y) = x2−1   ⟹   −1 ≤ x2−1 ≤ 1   ⟹   ∣x∣ ≤ 2\displaystyle Note: \ sin(2y) \ = \ x^{2}-1 \ \implies \ -1 \ \le \ x^{2}-1 \ \le \ 1 \ \implies \ |x| \ \le \ \sqrt 2

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

Ergo, f(2) = π4, f(−2) = π4, and f(0) = −π4, no inflection\displaystyle Ergo, \ f(\sqrt 2) \ = \ \frac{\pi}{4}, \ f(-\sqrt 2) \ = \ \frac{\pi}{4}, \ and \ f(0) \ = \ -\frac{\pi}{4}, \ no \ inflection

points, see graph.\displaystyle points, \ see \ graph.

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

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ddx[arcsin(u)] = u′(1−u2)\displaystyle \frac{d}{dx}[arcsin(u)] \ = \ \frac{u'}{\sqrt(1-u^{2})}

Now u = (x2−1), u′ = 2x and u2 = (x2−1)2\displaystyle Now \ u \ = \ (x^{2}-1), \ u' \ = \ 2x \ and \ u^{2} \ = \ (x^{2}-1)^{2}
 
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