anonymous
  • anonymous
Let y = f(x) be defined by the equation e^xy+ln(y^(2)) = x. Find dy/dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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IrishBoy123
  • IrishBoy123
this requires implicit differentiation. is there a specific problem? [if it were my problem, i would simplify by writing \(e^{xy}+\ln(y^{2}) = x\) as \(f(x,y) = e^{xy}+2\ln y - x =0 \) and noting that \(y' = -\dfrac{f_x}{f_y}\) administratively simpler. usually.
anonymous
  • anonymous
@IrishBoy123 I have an answer I'm just not sure how to go about solving this is there any chance you can show me the steps?
anonymous
  • anonymous
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IrishBoy123
  • IrishBoy123
for \(f(x,y) = e^{xy}+2\ln y - x =0\) you have \[f_x = \dfrac{\partial f}{\partial x} = y e^{xy} -1\] \[f_y = \dfrac{\partial f}{\partial y} = xe^{xy} + 2 .\frac{1}{y}\] \[\therefore y' = - \dfrac{y e^{xy} -1}{xe^{xy} + 2 \frac{1}{y}} \\ = - \dfrac{y (ye^{xy} -1)}{xye^{xy} + 2 }\] are you in that kinda territory?!?!? i just latexed this as i did it.....
anonymous
  • anonymous
@IrishBoy123 the answer i got from my professor was (1-ye^(xy))/(xe^(xy)+2/y)
IrishBoy123
  • IrishBoy123
.....which is the second last line of my post the important thing is: do you know how to do it 🤔
IrishBoy123
  • IrishBoy123
if you are learning implicit diff, you need to follow it through that way. so loads of product rule and chain rule.
anonymous
  • anonymous
@IrishBoy123 i understand the implicit part i just don't get how you incorporate chain or product
IrishBoy123
  • IrishBoy123
take it term by term you are doing this : \[\dfrac{d}{dx} \left( e^{xy}+\ln y^2 \right) = \dfrac{d}{dx}(x) \] so start with \[\dfrac{d}{dx} \left( e^{xy}\right)\]
anonymous
  • anonymous
A solution using the Total Derivative is attached.
1 Attachment
anonymous
  • anonymous
The Mathematica program was used to produce the solution above.
IrishBoy123
  • IrishBoy123
so, using the world's favourite letter [u] for calculus substitutions, we have \(\dfrac{d}{dx} \left( e^{xy}\right) = \dfrac{d}{dx} \left( e^{u}\right)\) where \(u = xy \; [= u(x,y)]\) and \( \dfrac{d}{dx} \left( e^{u}\right) = \dfrac{d}{du} \left( e^{u}\right) .\dfrac{du}{dx} = e^u .\dfrac{du}{dx}\) [chain rule [and fundamental property of e]] so now we need to find \(\dfrac{du}{dx}\).... and \(\dfrac{du}{dx} =\dfrac{d}{dx}(xy)\) that means we use the product rule, ie \(\dfrac{d}{dx}(xy) = x.\dfrac{d}{dx}(y) + \dfrac{d}{dx}(x).y\) there's a lot of work here but this is also a potentially awesome learning experience🍀

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