tepid flicker
Can someone explain 4 and maybe 3 too to me? I have an exam in 7 days and I just can't wrap my head around 4. I know I have to calculate y'(x) by isolating it and writing it using y(x) and x, but everywhere I look online I find different final answers to mine
shadow ice
If you get a 0/0 error in a limit you can use l’hopital rule
This states that the limit is equal to the limit of the derivative of the numerator divided by the derivative of denominator
Note that the derivative of arctan(x) = 1/(x^2+1)
granite portal
4 is using implicit differentiation
it isn't a differential equation
differentiate as usual for both X and y, but multiply the derivative of y by dy/dx and then isolate dy/dx
spice ore
Can someone explain 4 and maybe 3 too to me? I have an exam in 7 days and I just...
3√(x + y) + xy = y^2
We need to find dy/dx. Let's differentiate the expression.
(3/2)(dx + dy)/√(x + y) + ydx + xdy = 2ydy
Let's separate terms with dx and dy.
(2y - x - (3/2)/√(x + y))dy = ((3/2)/√(x + y) + y)dx
Then express dy/dx from there. The rest is easy.
3.
You can use l'Hospital, but i prefer not to do that.
Instead, let's remember the serues representations for arctangent and sine.
For x -> 0 we have:
arctan(x) = x - x^3/3 + x^5/5 + O(x^7)
sin(2x) = 2x - (2x)^3/6 + (2x)^5/120 + O(x^7) = 2x - (4/3)x^3 + (4/15)x^5 + O(x^7)
So:
3x - 3arctan(x) - x^3 = 3x - 3(x - x^3/3 + x^5/5 + O(x^7)) - x^3 = -(3/5)x^5 + O(x^7), x -> 0
6x - 3sin(2x) - 4x^3 = 6x - 3(2x - (4/3)x^3 + (4/15)x^5 + O(x^7)) - 4x^3 = -(4/5)x^5 + O(x^7), x -> 0
Thus:
(3x - 3arctan(x) - x^3)/(6x - 3sin(2x) - 4x^3) = (-(3/5)x^5 + O(x^7))/(-(4/5)x^5 + O(x^7)) = (3 + O(x^2))/(4 + O(x^2)) -> 3/4, x -> 0
Thus, the limit is equal to 3/4.
tepid flicker
Thank you very much! I did L'hospital rule but it was extremely lengthy and it would have not been efficient at an exam. Thanks again