I'm having trouble figuring out where to go with this implicit differentiation problem.

Problem: Find $\frac{dy}{dx}$ given that $\sin(x)=e^{-y\cos(x)}$

Here is how I start:

  • $d(\sin(x))=d(e^{-y\cos(x)})$
  • $dx\cos(x)=e^{-y\cos(x)}(-dy\cos(x)+y(\sin(x)))$

Here is where I get stuck, and I've tried a lot of different manipulations. I'm not sure how to isolate $\frac{dy}{dx}$. There's no way to factor or combine like terms -that I can see- that allows one to isolate $dy$, nor $dx$, but I know of course that I'm just missing something.

Solutions with identification of relevant principle requested please.

  • 1
    $\begingroup$ the second line of "how I start" should be $dx\cos x =e^{-y\cos(x)}(-dy\cos x +(dx) y\sin x)$ $\endgroup$ – John Joy Jun 11 '14 at 13:52
  • $\begingroup$ Why don't you just solve for $y$ and differentiate? $\endgroup$ – Christian Blatter Jun 11 '14 at 13:55
  • $\begingroup$ @JohnJoy - yep, it would appear I've been unfaithful to the chain rule. I can only hope in the future I'll be a better man to her. $\endgroup$ – Adam Jun 11 '14 at 23:28

Proceed like that: $d(e^{-y\cos x})=e^{-y\cos x}d(-y\cos x)=e^{-y\cos x}(y\sin x dx-\cos x dy)$;

$d\sin x=\cos x\,dx$

$\cos x\,dx=e^{-y\cos x}(y\sin x dx-\cos x dy)$

$$ (\cos x-ye^{-y\cos x}\sin x) dx=-e^{-y\cos x}\cos x dy $$

$$ \frac{dy}{dx}=\text{do it yourself} $$

The "relevant principle" is that $d(fg)=f dg +g df$.

  • $\begingroup$ Ahh, just that one dx- thanks very much for your help. $\endgroup$ – Adam Jun 11 '14 at 21:52

Just take the derivative of both sides of the equation $\sin x = e^{-y\cos x}$ $$d \sin x / dx = e^{-y\cos x}\cdot \frac{d(-y\cos x)}{dx}$$ $$\cos x = e^{-y\cos x}\cdot (-y'\cos x + y\sin x)$$ $$e^{y\cos x}\cos x = -y'\cos x + y\sin x$$ $$e^{y\cos x}\cos x - y\sin x= -y'\cos x$$ $$y' = (e^{y\cos x}\cos x - y\sin x)/(-\cos x) = -e^{y\cos x}+y\tan x$$


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