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[Solved] If X 2 + Xy + Y 3 = 1, Find The Value Of Y''' At The... | SolutionInn

zack May 06, 2021 No comments

What should be added to the product of (x^2 + xy - y^2 ) and (x^2 - xy + y^2 ) to get x^2 y^2 ?+xy+y= 1,find the value of y'''at the point where x= 1. The student who asked this found it Unhelpful.If xy + 9e^y = 9e, find the value of y'' at the point where x = 0. Okay so first I found the first derivative using implicit differentiation and I got: [itex]y'=\frac{-y} First, are we agreed that y = 1 for x = 0 ? (I want to be careful, since you only have one attempt left on your computer-problem system.)x2 - xy + y2 = 21 x2 + 2xy - 8y2 = 0. The absolute value of x in the second equation above gives two cases for the values of y Not all combinations of these x-value and y-values are solution points. Don't be sloppy; write the solution out correctly.x2+xy+y3=1. (if x is 1). 1×2+1×y+y×3=1. Give my ans thnx Nhi to popatlal ki shadi tumse hi hogi. the mean and median of 100 item's are 50 and 52 respectively it was later found that the value of the largest item is recorded as 100 instead of 110 c … ompute the corrected mean and median.

[Solved] If x 2 +xy+y 3 = 1, find the value ofy'''at the... | Course Hero

Ooops! That "C" can be a different value to the "C" just before. But they both mean "any constant", so let's call them C 1 and C 2 and then roll them into the C below by saying C=C 1 +C 2. Now we are going to find the function I(x, y). xcos(xy) + 12 e 2x + 13 y 3 = C. Done!xy^2 + 2xy = 8, then, at the point (1,2) y' is? algebra. The following identity can be used to find Pythagorean triples, where the expressions The line that is normal to the curve x^2=2xy-3y^2=0 at(1,1) intersects the curve at what other point? Please help. Thanks in advance. We have x2=2xy...Calculus. Find dy/dx x^2y^3-2xy=6x+y+1.Polynomials class 9. Polynomials class 9 maths important questions. if x+1/x=3 then find the value of x^3+1/x^3. if x+1/x=6 find x2+1/x2#polynomialsbyavisir...

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...Exponential, Logarithmic, And Inverse Trigonometric Functions 7 Techniques Of Integration 8 Further Applications Of Integration 9 Differential Equations 10 Parametric Equations And Polar Coordinates 11 Infinite Sequences And Series A Numbers, Inequalities, And Absolute Values B Coordinate Geometry...Click hereto get an answer to your question If 2x + y = 3 , xy = 1 , then the value of 8x^3 + y^3 is? The difference between two positive numbers is 4 and difference of their cubes is 316. Find their product.Originally Answered: If xy=2x+3y, how do I find the value of xy? Originally Answered: If xy=2x+3y find the value of xy? Do you mean what are the possible solutions in integers? Rewrite as y = 2x/(x-3) now x-3 divided into 2x leaves remainder 6 unless x-3 is a factor of 6Find solutions for your homework or get textbooks.If. x2 + xy + y3 = 1 Can you find the area of ABCD? what is the ratio of public health offices to people in Racine Wisconsin? In diagram, triangle RST is inscribed in circle O w diameter RT.

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Yes I'm conscious I posted this but I did earlier than adding these kind of main points and I can't wait for it to head off grasp... This is the question I will not seem to determine easy methods to do. If $x^2 + xy + y^3 = 1$, find the value of $y'''$ at the point where $x = 1$.

I need help with this because my instructor barely speaks English and doesn't answer questions.. Therefore I'm having a good looking time attempting the homework. What I believe I've been in a position to take action some distance is take the first by-product then setting $x=1$ $$ y' =\frac -2 y+3y^2 $$

Then looking to take the 2nd spinoff the usage of the quotient rule.. $$ y'' = \frac (y+3y^2)(0)-(-2)(y+3y^2)' (y+3y^2)^2 $$ $$ (y+3y^2)' = (1)\left(\fracdydx\right)+6y\left(\fracdydx\right) = \frac-2y+3y^2+\frac-12yy+3y^2 = \frac-2-12yy+3y^2$$ $$ y''=\frac0-(-2)\left(\frac-2-12yy+3y^2\proper)(y+3y^2)^2 = \frac -\frac 4+24y y+3y^2(y+3y^2)^2=-\frac 4+24y y+3y^2*\frac1(y+3y^2)^2$$ $$ y'' = \frac 4+24y (y+3y^2)^3 $$ At this point, I'm now not certain it this is right.. But I went and tried for the third with more quotient rule! $$y'''= \frac(y+3y^2)^3(24)\left(\fracdydx\proper)-(4+24y)\left((y+3y^2)^3\proper)'(y+3y^2)^6$$ Before I will finish that downside I want the spinoff of $(y+3y^2)^3$... I imagine for this I need to use chain Rule.if $f(x)=x^3$ then $f'(x)=3x^2$ If we then let $g(x) = y+3y^2$ then $g'(x) = 1+6y$ leaving us with... $ (y+3y^2)^2*(1+6y)$$ Plugging that again in to the leisure of the downside we get... $$y'''= \frac(y+3y^2)^3(24)\left(\fracdydx\right)-(4+24y)\left(3(y+3y^2)^2*(1+6y)\right)\left(\fracdydx\proper)(y+3y^2)^6$$ factoring out a $(y+3y^2)^2$ I am getting $$y'''=\frac(y+3y^2)^2\left(y+3y^2)(24)\left(\fracdydx\proper)-(4+24y)\left(3(1+6y)\proper)\left(\fracdydx\proper)\proper)(y+3y^2)^6$$ canceling out for... $$y'''=\frac\left(y+3y^2)(24)\left(\fracdydx\right)-(4+24y)\left(3(1+6y)\right)\left(\fracdydx\right)\right) (y+3y^2)^4$$ $$y'''=\frac24y(1+3y)\left(\fracdydx\proper)-12(36y^2+12y+1)\left(\fracdydx\proper) (y+3y^2)^4$$ $$y'''=\frac\frac-48y(1+3y)y+3y^2+\frac24(36y^2+12y+1)y+3y^2 (y+3y^2)^4=\frac\frac-48y(1+3y)+24(36y^2+12y+1)y+3y^2 (y+3y^2)^4$$ Then in spite of everything... $$y'''=\frac-48y(1+3y)+24(36y^2+12y+1)y+3y^2*\frac1(y+3y^2)^4$$ $$y'''=\frac-48y(1+3y)+24(36y^2+12y+1)(y+3y^2)^5 $$

asked Oct 13 '14 at 2:44

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You have $x^2 + xy + y^3 = 1$ and need $y'''(1)$. Differentiating, we've: $

{title}

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x+y+xy'+3y^2y' = 0$$ Again: $

{title}

{content}

+y'+y'+xy''+6y(y')^2+3y^2y'' = 0 \implies \\implies 2 + 2y'+6y(y')^2+(x+3y²)y'' = 0$$ Yet once more: $

{title}

{content}

y''+6(y')^3+12yy'y''+(1+6yy')y''+(1+3y^2)y''' = 0$$

If $x = 1$, then $x^2+xy+y^3 = 1$ offers $y+y^3 = 0 \implies y(y^2+1) = 0 \implies y = 0$. In the first step, we get $

{title}

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+y' = 0 \implies y' = -2$$ Now we sub. $x = 1, y = 0, y' = -2$ in the second step, to get: $

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-4+y'' = 0 \implies y'' = 2$$ Plugging all in the closing step: $-48+2+y''' = 0 \implies y''' = 42.$$

responded Oct 13 '14 at 3:03

Ivo TerekIvo Terek

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Differentiating $x^2 + xy + y^3 = 1$ gives $

{title}

{content}

x + xy' + y + 3y^2y' = 0$$ Differentiating again gives $

{title}

{content}

+ xy'' + 2y' + 6y(y')^2 + 3y^2 y'' = 0$$ A 3rd time offers $$xy''' + 3y'' + 6(y')^3 + 18 y y' y'' + 3y^2y''' = 0$$ In the best equation, settign $x = 1$ and $y = 0$ offers $

{title}

{content}

+ y'= 0$$ Which solves to $y' = -2$. Inserting into the 2nd equation gives $

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+ y'' - 4 = 0$$ This solves to $y'' = 2$. Putting into the 3rd equation provides $$y''' + 6 - 48 = 0$$ This solves to $y''' = 42$.

spoke back Oct 13 '14 at 3:01

ZarraxZarrax

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