Find (dy/dx) by implicit differentiation.?

use product rule

when you differentiate that equation according to x you get

(2xy^2+2y(dy/dx)x^2)+(siny+xcosy(dy/dx))=0

solve for dy/dx

2y(dy/dx)x^2)+xcosy(dy/dx)=-2xy^2-siny

dy/dx=(-2xy^2-siny)/(2yx^2+xcosy)

.

Take the spinoff of every time period in-place. Recall the chain rule around any y-time period offers you dy/dx accelerated after each y-time period. We'd like energy rule and consistent term rule. X^3 + y^four = 4 3x^2 + 4y^three(dy/dx) = zero remedy for dy/dx. 4y^3(dy/dx) = 3x^2 dy/dx = (3x^2) / (4y^three)

Do dy/dx to both sides

2x/y^2 + 2yx^2(dy/dx)/-2y^3 + siny + xcosy(dy/dx) = 0

(dy/dx)[(-x^2/y^2)+ xcosy] = -2x/y^2 - siny

dy/dx =[2x/y^2 -siny]/x^2/y^2 -xcosy] = [2x -y^2siny]/[x^2 -xy^2cosy]

x^2*y^2 + xsin(y) = 1

Differentiating implicitly:

2(xy)*(xy'+y) + xy'cos(y) + sin(y) = 0

y' = (-sin(y) - 2xy^2)/(2x^2y + xcos(y))

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Note:

  d(uv)

——– = u·v' + v·u'

  dx

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                                (x²y²) + (x siny) = 1

(x²•2yy' + y²•2x) + (x•cosy y' + siny•1 = 0

           2x²yy' + x cosy y' +2xy² + siny = 0

                               (2x²y + x cosy)y' = -(2xy² + siny)

                                                            -(2xy² + siny)

                                                    y' = ———————            ← ANSWER

                                                            2x²y + x cosy

Have a good one!

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