Christos
  • Christos
Implicit differentiation calculus, can you help me solve the first one please. https://www.dropbox.com/s/93h9x2h270cdhk8/Screenshot%202015-12-07%2016.14.45.png?dl=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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welshfella
  • welshfella
Treat y as a function of x and use the chain rule x^2 + y^2 = 100 2x + 2y y' = 0 solve for y'
Christos
  • Christos
-x/y
welshfella
  • welshfella
No 4 conatains 3xy^2 - here you use the product rule/chain rule.

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welshfella
  • welshfella
correct
Christos
  • Christos
why if I first solve for y and differentiate it gives different result ? @welshfella
welshfella
  • welshfella
- that way you get the derivative in terms of x . It will have the same value as the derivative in terms of x and y.
Christos
  • Christos
can you please explain to me what you just said ?
Christos
  • Christos
am I not looking for dy/dx in both cases ?
welshfella
  • welshfella
the derivative will be a different expression but there will only be x in it - no y's. yes
Christos
  • Christos
but they will be the same thing both of them, yes ?
welshfella
  • welshfella
they will both be the slope of a tangent to the curve at any point
Christos
  • Christos
same tangent , right?
Christos
  • Christos
ah i see
welshfella
  • welshfella
Its not always possible to find y in terms of x . That;s when implicit differentation is useful.
Christos
  • Christos
so implicit differentiation helps you find it in terms of x y
welshfella
  • welshfella
yes
Christos
  • Christos
thanks
welshfella
  • welshfella
yw
Christos
  • Christos
so just to sum up, the anwer to #4 is just -x/y ? no further steps ?
welshfella
  • welshfella
thats correct
Christos
  • Christos
ok
phi
  • phi
**why if I first solve for y and differentiate it gives different result ?*** You should get the same answer (though sometimes solving for y is not very practical) For example , Question 3-12, 3. \[ x^2+y^2= 100 \] by implicit differentiation you get \[ y' = -\frac{x}{y}\] if you solve for y: \[ y= \sqrt{100-x^2} \] and \[ y'= \frac{1}{2} \frac{-2x}{\sqrt{100-x^2}} \\ y'= - \frac{x}{\sqrt{100-x^2}} \] if you replace the denominator with the equivalent y you get \[ y'= -\frac{x}{y}\]
Christos
  • Christos
how would you replace the denominator with the equivalent y ? @phi
phi
  • phi
when you solved for y, you got \[ y= \sqrt{100-x^2} \] notice the right-hand side is the denominator in y' so you can put in y instead of \( \sqrt{100-x^2} \)
Zarkon
  • Zarkon
we should prob use \(y= \pm\sqrt{100-x^2}\) then \(\displaystyle y'=\pm\frac{1}{2}\frac{-2x}{\sqrt{100-x^2}}\) simplify... \[y'=\pm\frac{-x}{\sqrt{100-x^2}}=\frac{-x}{\pm\sqrt{100-x^2}}\] replace \(y\) \[y'=\frac{-x}{y}\]

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