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Study23
 2 years ago
Implicit Differentiation, please help!
\(\ \large cos(xy)=xe^x . \)
Study23
 2 years ago
Implicit Differentiation, please help! \(\ \large cos(xy)=xe^x . \)

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freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0Implicit 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 \[\cos\bigg(xy^2\bigg)\cdot\bigg( y^2 + x\cdot(2yy^{\prime})\bigg)\]

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0What I'm not understanding is that cos(xy) part. How do I take the derivative of that??

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0If you substituted u(x) for (xy) 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?

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry that should be \[sin(u(x))\cdot (u'(x))\]

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0So the two function we have here are cos x and xy? So we would use the chain rule?

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0yes. the only thing is that when you take the derivative of u(x) you are taking the derivative of "xy" 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.

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0crap, put a minus sign there in front of the y and y prime

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0So for the derivative of xy, with would be 1 y dy/dx ? Why doesn't the y become a one too?

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0no it would be 1 dy/dx you're instincts are correct

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{d}{dx}(xy) \to\frac{d}{dx}(x)+\frac{d}{dx}(y) \to 1+y' \]

Study23
 2 years ago
Best ResponseYou've already chosen the best response.0Okay, so what should the correct answer be? So I can check if I did this correctly??

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0So 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

freewilly922
 2 years ago
Best ResponseYou've already chosen the best response.0I'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. \[x+y' \sin(x)= y' + xy'\] \[x\sin(x)= y'(1 + x)\] \[\frac{xsin(x)}{1x}=y'\] notice for this problem you will not get a y prime on the RHS since there is no y variable in that function.
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