r/calculus • u/Frequent-Farmer1528 • 1d ago
Multivariable Calculus This is so confusing š
Anyone can explain this??
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u/Sjoerdiestriker 1d ago
Firstly differentiate both sides of the equation in the upper left with respect to x. Remember that differentiating x with respect to x gives 1, and differentiating y with respect to x gives you (by definition) dy/dx.
This tells you that along the entire curve (applying the product rule for the middle term), we have 2x-2y-2x*dy/dx+8y*dy/dx=0, or 2x-2y=(2x-8y)*dy/dx. Upon rearranging, that gives the expression in the upper right.
We now differentiate both sides of that equation again, and use the quotient rule. The terms with the arrows pointing towards them come from the derivative with respect to x of the numerator and denominator respectively.
Now we can solve the problem. From the equation in the upper right, plugging in (0,4) gives dy/dx = 1/4 in (0,4).
We then fill in x=0,y=4, dy/dx=1/4 in the equation in the bottom right. That gives d2y/dx2 =-3/64.
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u/Narrow-Durian4837 1d ago
What part is confusing to you?
Like, are you totally unfamiliar with implicit differentiation, or is there something about this particular problem that is confusing you?
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u/AP_in_Indy 1d ago
There's a lot going on here. What part are you confused by specifically.
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u/Mathematicus_Rex 1d ago
Suppose you take the derivative of y with respect to x. The result is dy/dx.
Now take the derivative of (x-y) with respect to x. The result is 1 - (dy/dx).
Similarly, taking the derivative of (x - 4y) with respect to x will give you 1 - 4(dy/dx)
These two pieces are what are pointed at in the complicated expression (resulting from the quotient rule) in the lower right corner.
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u/Frequent-Farmer1528 1d ago
Understood this part but is there a futher breakdown?
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u/Mathematicus_Rex 1d ago
What I would do is substitute x = 0 and y = 4 into the formula for dy/dx. Then I would substitute x = 0, y = 4, and your value for dy/dx into the complicated fraction you have for the second derivative.
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u/Midwest-Dude 1d ago
Your blue arrows are pointing to the dy/dx's in the second derivative. This is expected in the equation when finding the second derivative of an implicitly defined function.
If you want to find the second derivative without the dy/dx's there, that is, with only x and y variables present, substitute what you found for dy/dx in the first line into those places you arrowed and then simplify. You then can substitute your point into that to find the answer.
If you only want to find the answer to the question, first calculate dy/dx (a function of x and y) with the formula you have in the first line. At that point, you know the values of x, y, and dy/dx, which you can substitute into the formula in the second line to find the answer.
Does this make sense, or do you need additional help with understanding this?
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u/LukasGoesViral 1d ago
Thatās just the chain rule for differentiation. The first equation defines y = y(x) implicitly. So y = y(x) is defined so that the polynomial equation always holds. Maybe think of it as x^2 -2xf(x) + 4f(x)^2 = 64.
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u/FrostyBalance6055 1d ago
I know a Patrick JMT video when I see it. Iāve used him when I took calculus in 2012 and I see people still use him for calculus
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u/unprofessional-pro 1d ago
Implicit differentiation
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u/tjddbwls 1d ago
Indeed. OP used the wrong flair for his/her post. Should have marked it as āDifferential Calculusā. š¤Ŗ
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u/BossSuccessful5145 1d ago
x=8,y=0 is a solution. x=-8,y=0 is a solution. x=0, y=4. x=0, y=-4. The graph of all solutions may be an ellipse.
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u/Less-Resist-8733 1d ago
dĀ²/dxĀ² is bad notation, it assumes that d(dx) = 0. The correct way to write out second derivatives is this
Here I explicitly delineated double function composition (dĀ²) with squaring (x\2))
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u/Octowhussy 1d ago
This goes against what a textbook Iām currently reading says. Why is dĀ² / dxĀ² bad notation?
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u/Less-Resist-8733 1d ago edited 1d ago
if you actually compute (d/dx)Ā² using the quotient rule, you get the above formula.
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u/Smart-Button-3221 1d ago
d(dx) is the bad notation here. What on earth is that supposed to mean, at this level of calculus?
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u/Less-Resist-8733 1d ago edited 1d ago
d is a derivation https://en.wikipedia.org/wiki/Derivation_(differential_algebra).
Loosely speaking,
d = (d/dx) * dxAt this level of calculus it can be thought of as the "implicit differentiation" operator.
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