Question

Implicit Differentiation, please help!
\(\ \large cos(x-y)=xe^x . \)

Answer 1

Implicit D, is basically the chain rule without finishing.

In general you have something like
\[f(x) = (\text{stuff})^5\]
and
\[f'(x) =5(\text{stuff})^4 (\text{stuff})^{\prime}\]
Well you do the same thing here except when you have to do something like
\[\frac{d}{dx} y^2= 2y y'\]

So for example Implicit D of \[\sin\bigg(xy^2\bigg)\] give \[\cos\bigg(xy^2\bigg)\cdot\bigg( y^2 + x\cdot(2yy^{\prime})\bigg)\]

Answer 2

What I'm not understanding is that cos(x-y) part. How do I take the derivative of that??

Answer 3

If you substituted u(x) for (x-y) you would have
\[\frac{d}{dx} \cos(u(x))\]
which according to the chain rule would be
\[-\sin(u(x)\cdot (u'(x))\]
do you follow that?

Answer 4

Sorry that should be \[-sin(u(x))\cdot (u'(x))\]

Answer 5

So the two function we have here are cos x and x-y? So we would use the chain rule?

Answer 6

yes. the only thing is that when you take the derivative of u(x)
you are taking the derivative of "x-y" with respect to x.

So you would have
\[\frac{d}{dx}x + \frac{d}{dx}y\]
which becomes

\[1 + \frac{d}{dx}y\]

or

\[1 + y'\]

The fact that you don't finish taking the derivative of y, because you don't know what it is is the "unfinished" part of the chain rule.

Answer 7

crap, put a minus sign there in front of the y and y prime

Answer 8

So for the derivative of x-y, with would be 1- y dy/dx ? Why doesn't the y become a one too?

Answer 9

no it would be 1 -dy/dx
you're instincts are correct

Answer 10

\[\frac{d}{dx}(x-y) \to\frac{d}{dx}(x)+\frac{d}{dx}(-y) \to 1+y' \]

Answer 11

Okay, so what should the correct answer be? So I can check if I did this correctly??

Answer 12

So for an exponential function

\[\frac{d}{dx} e^{x^2y^3}\to e^{x^2y^3}\frac{d}{dx}(x^2y^3) \] and so on

Answer 13

Okay. Thanks!

Answer 14

I've been explicitly warned not to give answers to problems directly. The best I can do is give similar examples and hope that it helps.

The general process though is to implicitly differentiate both sides and then collect and isolate the y'.

so if you had
\[x+y' = \sin(x) + xy'\]
you would isolate the terms with y prime, factor y prime out and then divide by the non y prime factor.
\[x+y' -\sin(x)= -y' + xy'\]
\[x-\sin(x)= y'(-1 + x)\]

\[\frac{x-sin(x)}{1-x}=y'\]
notice for this problem you will not get a y prime on the RHS since there is no y variable in that function.