U = x log xy x 3 y 3 3xy = 1 By using total differentiation concept, \(du = \left( {\frac{{\partial u}}{{\partial x}}} \right)dx \left( {\frac{{\partial uEasy as pi (e) Unlock StepbyStep Natural Language Math InputFirst, differentiate with respect to x (use the Product Rule for the xy 2 term)
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(x+y)^3=x^3+y^3+3xy(x+y)
(x+y)^3=x^3+y^3+3xy(x+y)- Ex 25, 9 Verify (i) x3 y3 = (x y) (x2 – xy y2) LHS x3 y3 We know (x y)3 = x3 y3 3xy (x y) So, x3 y3 = (x y)3 – 3xy (x y) = (x y)3 – 3xy (x y) = (x y) (x y)2 – 3xy) Using (ab)2 = a2 b2 2ab = (x y) (x2 y2 2xy) – 3xy = (x y) (x2 y2 – xy) = (x y) (x2 xy y2) = RHS RHS (x y) (x2 – xy y2) Hence proved0 votes 1 answer Show that the set of curves intersect orthogonally x^3 – 3xy^2 = – 2 and 3x^2y – y^3 = 2 asked in Derivatives by RahulYadav (531k points)
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(a) x 2 y 2 2xy (b) x 2 y 2 – xy (c) xy 2 (d) 3xy polynomialsView Homework Help hw12soln from MATH 2940 at Cornell University 181 19) Let x3 y 3 = 3xy be the folium of Descartes (a) Show that the folium has aSee the answer See the answer See the answer done loading
Take the positive root since y(1) = 3 The restriction on x would be that 28−3x2 ≥ 0 Therefore, − s 28 3 < x < s 28 3 5 Problem 15 (xy2 bx2y)dx(xy)x2 dy = 0 First, for this to be exact M y = 2xy bx2 = 3x2 2xy = N x So b = 3 With this, find the solution to the DE f(x,y) = Z M dx = Z xy 23x2ydx = 1 2 x y2 x3y g(y) And solve #(xy)^3=(xy)(xy)(xy)# Expand the first two brackets #(xy)(xy)=x^2xyxyy^2# #rArr x^2y^22xy# Multiply the result by the last two brackets #(x^2y^22xy)(xy)=x^3x^2yxy^2y^32x^2y2xy^2# #rArr x^3y^33x^2y3xy^2#Calculus Find dy/dx x^3y^3=3xy^2 x3 y3 = 3xy2 x 3 y 3 = 3 x y 2 Differentiate both sides of the equation d dx (x3 y3) = d dx (3xy2) d d x ( x 3 y 3) = d d x ( 3 x y 2) Differentiate the left side of the equation Tap for more steps Differentiate
Consider the function f(x;y) = 3 p x2y Show that the directional derivative exists at the origin (by letting ~u= hcos( );sin( )iand using the de nition), BUT, fis not di erentiable at the origin (because if it were, we could use rf~uto compute D ~uf) SOLUTION Compute the directional derivative at the origin by using the de nition D uf(0;0X^3y^3=3xy WolframAlpha Area of a circle? User Simplify 2(x y) 3(x y)2xy 3xy 5xy 5x 5y Weegy 2(x y) 3(x y) 2x 2y 3x 3y = 5x 5y paralPoints User Simplify 15 6(x 1)6x 15 6x 21 6x 16 User Simplify 5(MN 1) 55MN 5MN 4 5MN 4 Weegy 5(MN 1) 5 = 5MN 5MN 5 5 = 5MN User Simplify 8(xy 8) 18xy 9 8xy 65 8xy 73
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Fall 13 S Jamshidi 4 x4 y4 z4 =1 If x,y,z are nonzero, then we can consider Therefore, we have the following equations 1 1=2x2 2 1=2y2 3 1=2z2 4 x4 y4 z4 =1 Remember, we can only make this simplification if all the variables are nonzero!Factor out the Greatest Common Factor (GCF), '3xy' 3xy(y x) = 0 Ignore the factor 3 Subproblem 1 Set the factor 'xy' equal to zero and attempt to solve Simplifying xy = 0 Solving xy = 0 Move all terms containing x to the left, all other terms to the right Simplifying xy = 0 The solution to this equation could not be determinedSo we set it = 0 i*sqrt(3)(x1) = 0 x1 = 0 x = 1 Then substituting that y = y = y = y = 1 So we end up with all solutions {(x,y) x y = 1} plus the one solution (x,y) = (1,1) Now if you were asking about xy instead of xy, the answer would be xy is always either 1 or 2 Are you sure you didn't make a typo and you were asking about xy and not xy?
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Phân tích đa thức x^3y^3z^33xyz thành nhân tử Phân tích đa thức thành nhân tử x3 y3 z3 −3xyz x 3 y 3 z 3 − 3 x y z ( Làm rõ từng bước đừng làm tắt giúp mình với nhé ) Theo dõi Vi phạm Toán 8 Bài 6 Trắc nghiệm Toán 8 Bài 6 Giải bài tập Toán 8 Bài 6 ADSENSEIt is clear that when $x=y$ we have $x^3y^3=0$ Then use long division to divide $x^3y^3$ by $xy$ and the result will be the equation on the right Another way would be to write $$\left(\frac{x}{y}\right)^3 1$$ Now we wish to find the zeros of this polynomialX y is a binomial in which x and y are two terms In mathematics, the cube of sum of two terms is expressed as the cube of binomial x y It is read as x plus y whole cube It is mainly used in mathematics as a formula for expanding cube of sum of any two terms in their terms ( x y) 3 = x 3 y 3 3 x 2 y 3 x y 2
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F(x)=x³y³3xy f'x=3x²3y=0, 得y=x² f'y=3y²3x=0，代入y得：x^4x=0, 得x=0, 1 得驻点(0, 0) (1, 1) A=f"xx=6x B=f"xy=3 C=f"yy=6y 在(0, 0), A=C=0, B²AC=9>0, 所以这不是极值点；(x y) 3 = x 3 3x 2 y 3xy 2 y 3 Example (1 a 2 ) 3 = 1 3 31 2 a 2 31(a 2 ) 2 (a 2 ) 3 = 1 3a 2 3a 4 a 6 (x y z) 2 = x 2 y 2 z 2 2xy 2xz 2yzAnswer to Find the general solution of the equation x(3xy 4y^3 6)dx (x^3 6x^2 y^2 1)dy = 0 By signing up, you'll get thousands of
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SolutionShow Solution Elaborating x 3 y 3 using identity a 3 b 3 = (a b) (a 2 ab b 2 ) = x ( x y) (x 2 xy y 2 ) 3xy (x y ) Taking common x ( x y ) in both the terms = x ( x y) {x 2 xy y 2 3y} ∴ x (x 3 y 3 ) 3xy ( x y) = x ( x y ) (x 2 xy y 2 3y) If a=0, the function becomes x 3 y 3, which does not have any maximum or minimum and has a saddle point at (x,y)=(0,0) Therefore, we conclude as under 1 If a>0, the function does not have a maximum but has a local minimum at (a,a) having a value −a 3 2 If aSection 31 SecondOrderLinear Equations 311 Verify that the functions y 1 and y 2 given below are solutions to the secondorder ODE also given below Then, find a particular solution of the form y = c 1y 1 c 2y 2 that satisfies the given initial conditions Primes denote derivatives with respect to
X3 Y3 X Y X2 Xy Y2
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SOLUTION 15 Since the equation x 2 xy y 2 = 3 represents an ellipse, the largest and smallest values of y will occur at the highest and lowest points of the ellipse This is where tangent lines to the graph are horizontal, ie, where the first derivative y'=0Answer to Solved 8x3−6x^2y3xy^2/2 y^3/8=(2x−y/2)^3 This problem has been solved! Tính giá trị của biểu thức P=x^3y^33xy biết xy=1 Cho xy=1 Tính giá trị của a) P = x3 y3 3xy P = x 3 y 3 3 x y b) Q = x3 y3 3xy(x2 y2) 6x2y2 (x y) Q = x 3 y 3 3 x y ( x 2 y 2) 6 x 2 y 2 ( x y) Theo dõi Vi phạm Toán 8 Bài 3 Trắc nghiệm Toán 8 Bài 3 Giải bài tập Toán 8 Bài 3
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Find the critical points of the function {eq}f(x,y) = 4 x^3 y^3 3xy {/eq} and classify them as local maximum or minimum or saddle points Ex 25, 13 If x y z = 0, show that x3 y3 z3 = 3xyz We know that x3 y3 z3 3xyz = (x y z) (x2 y2 z2 xy yz zx) Putting x y z = 0, x3 y3 z3 3xyz = (0) (x2 y2 z2 xy yz zx) x3 y3 z3 3xyz = 0 x3 y3 z3 = 3xyz Hence proThe two curves x^3 – 3xy^2 2 = 0 and 3x^2y – y^3 = 2 asked in Mathematics by AsutoshSahni (528k points) application of derivative;
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If x and y are integers, then we can let y = x − k, where k is an integer Then the equation can be written as a quadratic in x k ( x 2 x ( x − k) ( x − k) 2) − x ( x − k) − 61 = 0 The solutions to this are x = k ( 3 k − 1) ± ( 1 − 3 k) ( k 3 k 2 − 244) 2 ( 3 k − 1) The term under the square root is only positive{(x,y) x y ≥ 1}, which is the region above the line y = 1 − x See figure above, right To compute the probability, we double integrate the joint density over this subset of the support set P(X Y ≥ 1) = Z 1 0 Z 2 1−x (x2 xy 3)dydx = 65 72 (c) We compute the marginal pdfs fX(x) = Z ∞ −∞ f(x,y)dy = ˆR 2 0 (x 2 xy 3)dySubject to the constraint 2x2 (y 1)2 18 Solution We check for the critical points in the interior f x = 2x;f y = 2(y1) =)(0;
Solved Find All Critical Points Of F X Y X3 Y3 3xy 4 Using The Discriminant And Second Derivative Test To Identify Each As A Local Maximum Local Minimum Or Saddle Point
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Get stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us! Which of the following is a factor of (x y) 3 – (x 3 y 3 )?Given, (x−y 3)dx3xy 2dy=0 Divide above equation by xdx 3y 2dxdy − xy 3 =−1 −(1) let y 3=t
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(xyz)^3 (x y z)(x y z)(x y z) We multiply using the FOIL Method x * x = x^2 x * y = xy x * z = xz y * x = xy y * y = y^2 y * z = yz z * x = xz z * y = yz z * z = z^2 x^3 y^3 z^3 3x^2y 3xy^2 3x^2z 3z^2x 3y^2z 3z^2y 6xyz Lennox Obuong Algebra StudentClick here👆to get an answer to your question ️ If x^3 y^3 = 3axy , find dydxUse implicit differentiation to find the largest yvalue in the loop of the Folium of Descartes, which is given by x3 y3 − 3xy = 0 First, we do the implicit derivative to simplify our equation Because maxima/minima occur when f' (x)=0, we take the implicit derivative and set it equal to 0
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Answer k=3 Stepbystep explanation we know that A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form or In this problem we have therefore The constant of proportionality is k=3 quarterfreelp and 26 more users found this answer helpful (xy)^3 (yz)3 (zx)^3 = 3(xy)(yz)(zx) That is it no constraints etc It mentions "This can be done by expanding out the brackets, but there is a more elegant solution" Homework Equations The Attempt at a Solution First of all this only seems to hold in special cases as I have substituted random values for x,y and z and they do not agreeAnswer (1 of 4) The graph of x^3y^3=3axy is known as the Folium of Descartes It cannot be factored Further, y is not a function of x, because the graph does does not
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We don't have to!Use the distributive property to multiply x y by x 2 − x y y 2 and combine like terms x^ {3}y^ {3}=x^ {3}y^ {3} x 3 y 3 = x 3 y 3 Subtract y^ {3} from both sides Subtract y 3 from both sides x^ {3}y^ {3}y^ {3}=x^ {3} x 3 y 3 − y 3 = x 3 Combine y^ {3} and y^ {3} to get 0The correct option is D Cut at right anglesx^3 3xy^2 2 = 0(1) 3x^2y y^3 2 = 0(2) On differentiating equations (1) and (2) wrt x, we obtain ( dy/
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The function F(x, y) = x 3 y 3 3xy 4 First order partial derivatives f x (x, y) = 3x 2 3y f y (x, y) = 3y 2 3x Second order partial derivatives F xx (x, y) = 6x F yy (x, y) = 6y F xy (x, y) = 3 Step 2 The critical points satisfy the equations f x (x,y) = 0 and f y (x,y) = 0 So solve the following equations F x = 0 and F ySOLUTION 1 Begin with x 3 y 3 = 4 Differentiate both sides of the equation, getting D ( x 3 y 3) = D ( 4 ) , D ( x 3) D ( y 3) = D ( 4 ) , (Remember to use the chain rule on D ( y 3) ) 3x 2 3y 2 y' = 0 , so that (Now solve for y' ) 3y 2 y' = 3x 2, and Click HERE to return to the list of problems SOLUTION 2 Begin with (xy) 2 = x y 1 Differentiate both sides1) is a critical point The second derivative test f xx = 2;f yy = 2;f xy = 0 shows this a local minimum with
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Ejercicios EDO's de primer orden 3 1 y3 dy = dx x2 Z y−3 dy = Z x−2 dx, 1 −2 y−2 = −x−1 c 1, −1 2y2 −1 x c 1, 1 y2 2 x c, c = −2c 1 Solución implícita 1 y2 2xc x Solución explícita y = ±Y = −3/4 x 25/4 Another Example Sometimes the implicit way works where the explicit way is hard or impossible Example 10x 4 − 18xy 2 10y 3 = 48 How do we solve for y?The two curves `x^(3)3xy^(2)2=0 and 3x^(2)yy^(3)=2`Welcome to Doubtnut Doubtnut is World's Biggest Platform for Video Solutions of Physics, Chemistry, Ma
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Factor x^3y^3 x3 − y3 x 3 y 3 Since both terms are perfect cubes, factor using the difference of cubes formula, a3 −b3 = (a−b)(a2 abb2) a 3 b 3 = ( a b) ( a 2 a b b 2) where a = x a = x and b = y b = y (x−y)(x2 xyy2) ( x y) ( x 2 x y y 2)
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