Use IMPLICIT differentiation to get dy/dx for xy2 + y + x2?

Question

could someone please help me with how to set it up and get the right answer thanks:)

Answer 1

Assume an equation:-
d/dx ( x y ² + y + x ² ) = 0
y² + 2y (dy/dx)(x) + dy/dx + 2x = 0
y² + 2x + (2xy + 1) (dy/dx)
dy/dx = - (y ² + 2x) / (2xy + 1)

Answer 2

one million. i'm assuming that it incredibly is xy^2 - 2x^2 -2y = -38 Then, differentiating with admire to x, y^2 +2xy (dy/dx) - 4x -2 (dy/dx) = 0 or (2xy -2) (dy/dx) + y^2-4x = 0 or dy/dx = (4x-y^2)/(2xy-2) Plugging in x=-4 and y=one million it incredibly is dy/dx = (-sixteen-one million)/(-8-2) = (-17)/(-10) = 17/10 2. int (from x=0 to x=2) (x^2+bx) dx = [ x^3/3+bx^2/2 ] from x=0 to x=2 = (8/3+2b) - (0+0) = 8/3 + 2b it incredibly is = 10 whilst 8/3 + 2b = 10 2b = 10-8/3 = 22/3 b = eleven/3 3. This one is complicated to make sure. Please use "^" to point exponents. If that's I = quintessential { x*(one million+x^2)^4 } dx Make the exchange of variables y=one million+x^2. Then dy = 2x dx and that i = (one million/2) quintessential { y^4 } dy = (one million/2) y^5/5 = y^5/10 Substituting returned in, y=one million+x^2, I = (one million/10) (one million+x^2)^5 4. you're able to do this one by way of areas or by way of substitution. as quickly as you have completed a number of those, you style of get the image approximately the way it works. I = quintessential { x ln x -one million } dx = quintessential {x ln x} dx - x do this one by way of areas. enable u = ln x, dv = x dx. So, v = x^2/2 and du = (one million/x) dx and that i = uv - quintessential {v} du - x = (one million/2) x^2 ln x - quintessential (one million/2) x^2 (one million/x) dx - x = (one million/2) x^2 ln x - (one million/2) quintessential x dx - x = (one million/2) x^2 ln x - (one million/4) x^2 - x To double verify - in simple terms differentiate! by-product = x ln x + (one million/2) x^2*(one million/x) - (2/4) x - one million = x ln x + (one million/2) x - (one million/2) x - one million = x ln x - one million as wanted 5. replace y = g(x), so dy = g'(x) dx and eight * quintessential { [g(x)]^3 g'(x) } dx = 8 * quintessential { y^3 } dy = 8 * y^4/4 = 2*y^4 = 2*[g(x)]^4 additionally, please remember that to any antiderivative, you may constantly upload an arbitrary consistent. =

Answer 3

take differentiation from each part one time for x and one time for y.
so for xy2+y+x2 is: y2+2xydy/dx+dy/dx+2x=0 so
dy/dx(2xy+1)=-(2x+y2) so dy/dx=-(2x+y2)/(2xy+1)