# Implicit Differentiation of a Trignometric Equation [Multiple Choice]

• Nov 7th 2009, 05:27 PM
StarlitxSunshine
Implicit Differentiation of a Trignometric Equation [Multiple Choice]
If $cos (xy) = x$, then $\frac{dy}{dx}=$

1. (A) $\frac{-csc (xy) - y}{x}$
2. (B) $\frac{csc (xy) - 1}{x}$
3. (C) $-csc (xy)$
4. (D) $\frac{-csc (xy)}{x}$
5. (A) $-csc (xy) - 1$
• Nov 7th 2009, 05:31 PM
VonNemo19
Quote:

Originally Posted by StarlitxSunshine
If $cos (xy) = x$, then $\frac{dy}{dx}=$

1. (A) $\frac{-csc (xy) - y}{x}$
2. (B) $\frac{csc (xy) - 1}{x}$
3. (C) $-csc (xy)$
4. (D) $\frac{-csc (xy)}{x}$
5. (A) $-csc (xy) - 1$

I give up...Which one is it?

(Rofl)

What's the problem, what don't you understand about the question? where are you stuck? Help me help you.(Wink)
• Nov 7th 2009, 05:33 PM
apcalculus
Quote:

Originally Posted by StarlitxSunshine
If $cos (xy) = x$, then $\frac{dy}{dx}=$

1. (A) $\frac{-csc (xy) - y}{x}$
2. (B) $\frac{csc (xy) - 1}{x}$
3. (C) $-csc (xy)$
4. (D) $\frac{-csc (xy)}{x}$
5. (A) $-csc (xy) - 1$

Implicit differentiation gives
$-\sin(xy) (y + x y') = 1$

Solve for y':

$y' = \frac{\frac{-1}{sin(xy)}-y}{x}=\frac{-csc(xy)-y}{x}$
• Nov 7th 2009, 05:39 PM
StarlitxSunshine
Quote:

Originally Posted by VonNemo19
I give up...Which one is it?

(Rofl)

What's the problem, what don't you understand about the question? where are you stuck? Help me help you.(Wink)

>_< I know how you feel xD

I tried implicit differentiation, but I don't think I'm doing it right. I've never done a question where the sin function has two variables (both x and y) in it, so I'm not really sure how that works. And moreover, I can't seem to get past the first step of the implicit differentiation because I don't know what to do next. :33
• Nov 7th 2009, 05:51 PM
VonNemo19
Quote:

Originally Posted by StarlitxSunshine
>_< I know how you feel xD

I tried implicit differentiation, but I don't think I'm doing it right. I've never done a question where the sin function has two variables (both x and y) in it, so I'm not really sure how that works. And moreover, I can't seem to get past the first step of the implicit differentiation because I don't know what to do next. :33

"implicit" means that it is "implied" that $y$ is a function of $x$. This means that when $x$ and $y$ are grouped together as a product, quotient, etc... it must be understood that one must employ the neccesary and appropriate method of differetiating.

For example, consider the function

$f(x)=y=x^2+x$

Taking the derivative is simple enough

$f'(x)=\frac{dy}{dx}=2x+1$

Well, what if we had simply rewrote before differentiating, and subtracted $y$ from both sides

$0=x^2+x-y$

No problem, $y$ is still a function of $x$, so the derivative "with respect to x" is

$0=2x+1-\frac{dy}{dx}$

And adding, we are back to where we started

$\frac{dy}{dx}=2x+1$.

In this case, it was very easy to solve for $y$, but sometimes - as in the problem you have provided - solving for $y$ is tedious, and in some cases, impossible. So, we understand $y$ to be an "implied" function of $x$and we differentiate.

Bye!