anonymous
  • anonymous
Determine the slope of the graph of x2 = ln(xy) at the point (1, e) Do I start with implicit differentiation? Or is there another way? I do not know what to do.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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xMissAlyCatx
  • xMissAlyCatx
@hero please help :)
Hero
  • Hero
Try isolating y first.
Hero
  • Hero
Afterwards, find y'

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benlindquist
  • benlindquist
so many people
anonymous
  • anonymous
\[y=\frac{ 1 }{ x ^{3} }\] @Hero So far so good?
SolomonZelman
  • SolomonZelman
You can isolate the y, and as I did that I obtained \(\large\color{#000000 }{ \displaystyle y(x)=e^{x^2}/x }\) (Note, if you want to integrate that, then top and bottom •x, and then set u=x\(^2\),)
SolomonZelman
  • SolomonZelman
9(No, you don't need to integrate this, this is not about integration at all, but just in case you wondered about it))
SolomonZelman
  • SolomonZelman
If you want, you could do implicit differentiation as well. \(\large\color{#000000 }{ \displaystyle x^2 = \ln(xy) }\) \(\large\color{#000000 }{ \displaystyle x^2 = \ln(x)+\ln(y) }\) then, differentiate each component. (where the derivative of ln(y) deserves a y')
anonymous
  • anonymous
Thanks!
SolomonZelman
  • SolomonZelman
Just in case, (to nail the concept if it's not nailed already). \(\large\color{#000000 }{ \displaystyle \frac{ d}{dx}\left[~\ln f(x)~\right]=\frac{1}{f(x)}\times f'(x)=\frac{f'(x)}{f(x)} }\) (This is when you take an "ln" of a function; via the chain rule) And that above is exactly the equivalent of \(\large\color{#000000 }{ \displaystyle \frac{ d}{dx}\left[~\ln y~\right]=\frac{1}{y}\times y'=\frac{y'}{y} }\) Because y is a function of x (just as f(x) is), and therefore, just as f(x) it desrves a chain rule. ((In case you might have been wondering where this y' always comes from; hope I addressed this question))

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