Implicit Differentiation of a Trignometric Equation [Multiple Choice]

**StarlitxSunshine** · November 7th 2009, 04:27 PM

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

(A) $\frac{-csc (xy) - y}{x}$
(B) $\frac{csc (xy) - 1}{x}$
(C) $-csc (xy)$
(D) $\frac{-csc (xy)}{x}$
(A) $-csc (xy) - 1$

**VonNemo19** · November 7th 2009, 04:31 PM

Originally Posted by StarlitxSunshine

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

(A) $\frac{-csc (xy) - y}{x}$
(B) $\frac{csc (xy) - 1}{x}$
(C) $-csc (xy)$
(D) $\frac{-csc (xy)}{x}$
(A) $-csc (xy) - 1$

I give up...Which one is it?

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

**apcalculus** · November 7th 2009, 04:33 PM

Originally Posted by StarlitxSunshine

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

(A) $\frac{-csc (xy) - y}{x}$
(B) $\frac{csc (xy) - 1}{x}$
(C) $-csc (xy)$
(D) $\frac{-csc (xy)}{x}$
(A) $-csc (xy) - 1$

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

Solve for y':

$y' = \frac{\frac{-1}{sin(xy)}-y}{x}=\frac{-csc(xy)-y}{x}$

**StarlitxSunshine** · November 7th 2009, 04:39 PM

Originally Posted by VonNemo19

I give up...Which one is it?

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

>_< 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

**VonNemo19** · November 7th 2009, 04:51 PM

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!

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