tmagloire1
  • tmagloire1
Can anyone help me on this problem regarding implicit differentiation? Thanks http://prntscr.com/7u700y
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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DanJS
  • DanJS
when differentiating y, remember to use the chain rule...
DanJS
  • DanJS
For example, \[\frac{ d }{ dx }[y^3] = \frac{ d }{ dy }[y^3]*\frac{ dy }{ dx }\]
DanJS
  • DanJS
Basic chain rule for this \[\frac{ d }{ dx } = \frac{ d }{ dy }*\frac{ dy }{ dx }\]

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More answers

DanJS
  • DanJS
Following the above example, \[\frac{ d }{ dx }[y^3] = 3y^2*\frac{ dy }{ dx }\]
DanJS
  • DanJS
Do that for the y terms, and differentiate the x terms normally, then solve for dy/dx
tmagloire1
  • tmagloire1
Okay I'll try it and tell you what I get.
DanJS
  • DanJS
k
tmagloire1
  • tmagloire1
Okay I am actually still very confused. Am i supposed to find the second derivative of 5x^2 + y4 = −9 and then use implicit differentiation?
DanJS
  • DanJS
Yes you want to find the second derivative of the given thing with respect to X, but to take the derivative w.r.t. X of the Y term, you have to use that chain rule... here ill type it out
DanJS
  • DanJS
\[\frac{ d }{ dx }[5x^2]+\frac{ d }{ dx }y^4 = \frac{ d }{ dx }[-9]\] \[\frac{ d }{ dx }[5x^2]+\frac{ d }{ dy }y^4*\frac{ dy }{ dx } = \frac{ d }{ dx }[9]\]
DanJS
  • DanJS
now take derivative of y term with respect to y like normal, but you have to put the y' on there too from the chain rule.
DanJS
  • DanJS
\[\frac{ d }{ dx }[5x^2]+\frac{ d }{ dy }y^4*\frac{ dy }{ dx } = \frac{ d }{ dx }[-9]\] \[10x^2+4y^3*\frac{ dy }{ dx } = 0\]
DanJS
  • DanJS
Solve that for dy/dx and that is your first derivative...
DanJS
  • DanJS
Or just remember, if you are taking the derivative of a Y term, tag on a dy/dx next to it ...basically
tmagloire1
  • tmagloire1
Okay for the first derivative I got : dy/dx= -10x^2 / 4y^3
tmagloire1
  • tmagloire1
And then take the derivative again?
DanJS
  • DanJS
So you get, \[\frac{ dy }{ dx } = \frac{ -10x^2 }{ 4y^3 }\] right
DanJS
  • DanJS
This time you need the product or the quotient rule and the chain rule...
DanJS
  • DanJS
I can start typing it out...
tmagloire1
  • tmagloire1
Im using the quotient rule
DanJS
  • DanJS
\[\frac{ d }{ dx }\frac{ dy }{ dx } = \frac{ d^2y }{ dx^2 }\] k, ill start typing it
tmagloire1
  • tmagloire1
I got : -80xy^3 +120x^2y^2 / (4y^3)^2
DanJS
  • DanJS
\[\frac{ d^2y }{ dx^2 } = \frac{ d }{ dx}\frac{ -5x^2 }{ 2y^3 }\]
tmagloire1
  • tmagloire1
Oh I think I took the derivative of the wrong thing?
tmagloire1
  • tmagloire1
So it would be -20 when you plug in x and y
DanJS
  • DanJS
\[\frac{ d^2y }{ dx^2 }=\frac{ 2y^2*-10x - [-5x*6y^2*\frac{ dy }{ dx } ]}{ 4y^6 }\]
DanJS
  • DanJS
That is the quotient rule , using the chain rule on the y term in the numorator.
tmagloire1
  • tmagloire1
So am I supposed to plug in x and y into which equation?
DanJS
  • DanJS
yes, but notice the second derivative has a term of the first derivative in the numerator... but you know what dy/dx is , so sub that in first
DanJS
  • DanJS
\[\frac{ dy }{ dx } = \frac{ -5x^2 }{ 2y^3 }\] reduced fraction... Sub that in for dyy/dx in the second derivative then plug in the x and y and solve...
DanJS
  • DanJS
Less writing overall if you want to use y' and y'' instead of dy/dx things
DanJS
  • DanJS
goodluck, i dont want to calculate it.
Jhannybean
  • Jhannybean
\[\large \color{blue}{\checkmark}\]
tmagloire1
  • tmagloire1
I got -310
tmagloire1
  • tmagloire1
Thanks for all the help!
DanJS
  • DanJS
yw, just get that Leibanitz notation dy/dx crap down, it makes the chain rule and this stuff easier i think .. goodluck
tmagloire1
  • tmagloire1
Okay I will try using it more.

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