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

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

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

- schrodinger

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

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 \[\cos\bigg(xy^2\bigg)\cdot\bigg( y^2 + x\cdot(2yy^{\prime})\bigg)\]

- anonymous

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

- anonymous

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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

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

- anonymous

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

- anonymous

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.

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Okay. Thanks!

- anonymous

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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.