Finding second derivative of sin y+cos y=x in terms of x

[sooyong94]

Homework Statement

Find the second derivative of sin y+cos y=x, giving your answer in terms of x.

Homework Equations

Implicit derivatives

The Attempt at a Solution

$\sin y+\cos y=x$
$\cos y \frac{dy}{dx} -\sin y \frac{dy}{dx}=1$
$\frac{dy}{dx}=\frac{1}{\cos y-\sin y}$

Now, do I use the chain rule by rewriting them as $(\cos y-\sin y)^{-1}$, or use the quotient rule instead?

 

[Mark44]

Homework Statement

Find the second derivative of sin y+cos y=x, giving your answer in terms of x.

Homework Equations

Implicit derivatives

The Attempt at a Solution

$\sin y+\cos y=x$
$\cos y \frac{dy}{dx} -\sin y \frac{dy}{dx}=1$
$\frac{dy}{dx}=\frac{1}{\cos y-\sin y}$

Now, do I use the chain rule by rewriting them as $(\cos y-\sin y)^{-1}$, or use the quotient rule instead?

Either one. The chain rule might be a little easier.

 

[sooyong94]

Now that gives:
$\frac{d^{2}y}{dx^{2}}=-(\cos y-\sin y)^{-2} (-\sin y \frac{dy}{dx} -\cos y \frac{dy}{dx})$

 

[Curious3141]

Now that gives:
$\frac{d^{2}y}{dx^{2}}=-(\cos y-\sin y)^{-2} (-\sin y \frac{dy}{dx} -\cos y \frac{dy}{dx})$

That should simplify to $\frac{d^2y}{dx^2} = x(\frac{dy}{dx})^3$ if you do the proper subs. Actually, it would've been much easier to leave everything on the LHS, and just use product rule and chain rule to get to that.

But it still doesn't help to put everything in terms of x. Honestly, I cannot see how you can progress from that stage.

I think you're much better off trying to find y as a function of x, then differentiating twice. This gives a quick answer.

To do that, think of how you would put $a\sin x + b\cos x$ into the form $c\sin(x + \theta)$.

EDIT: There is a way to continue from $\frac{d^2y}{dx^2} = x(\frac{dy}{dx})^3$. But you have to square the implicit equations for x and the first derivative and add them together to eliminate the y term (apply a trig identity as well). Seems like a bit more effort than just recasting y in terms of x from the start.

 

[sooyong94]

Well, the form $c\sin(x+\theta)$ worked nicely. :P