anonymous
  • anonymous
Implicit Differentiation, please help! \(\ \large cos(x-y)=xe^x . \)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
What I'm not understanding is that cos(x-y) part. How do I take the derivative of that??
anonymous
  • 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
  • anonymous
Sorry that should be \[-sin(u(x))\cdot (u'(x))\]
anonymous
  • anonymous
So the two function we have here are cos x and x-y? So we would use the chain rule?
anonymous
  • 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
  • anonymous
crap, put a minus sign there in front of the y and y prime
anonymous
  • 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
  • anonymous
no it would be 1 -dy/dx you're instincts are correct
anonymous
  • anonymous
\[\frac{d}{dx}(x-y) \to\frac{d}{dx}(x)+\frac{d}{dx}(-y) \to 1+y' \]
anonymous
  • anonymous
Okay, so what should the correct answer be? So I can check if I did this correctly??
anonymous
  • 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
  • anonymous
Okay. Thanks!
anonymous
  • 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.