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If 
-y+x+y^(2)+5y^(3)+3x^(2)=0 then find 
(dy)/(dx) in terms of 
x and 
y.
Answer: 
(dy)/(dx)=

If y+x+y2+5y3+3x2=0 -y+x+y^{2}+5 y^{3}+3 x^{2}=0 then find dydx \frac{d y}{d x} in terms of x x and y y .\newlineAnswer: dydx= \frac{d y}{d x}=

Full solution

Q. If y+x+y2+5y3+3x2=0 -y+x+y^{2}+5 y^{3}+3 x^{2}=0 then find dydx \frac{d y}{d x} in terms of x x and y y .\newlineAnswer: dydx= \frac{d y}{d x}=
  1. Given Equation: We are given the equation y+x+y2+5y3+3x2=0-y + x + y^{2} + 5y^{3} + 3x^{2} = 0 and we need to find the derivative of yy with respect to xx, which is dydx\frac{dy}{dx}. To do this, we will use implicit differentiation, which involves taking the derivative of both sides of the equation with respect to xx, while treating yy as a function of xx.
  2. Implicit Differentiation: First, we differentiate each term of the equation with respect to xx. For the terms involving only xx, we use the standard rules of differentiation. For the terms involving yy, we use the chain rule, remembering to multiply by dydx\frac{dy}{dx} since yy is a function of xx. The derivative of y-y with respect to xx is 1dydx-1\frac{dy}{dx}. The derivative of xx with respect to xx is xx11. The derivative of xx22 with respect to xx is xx44. The derivative of xx55 with respect to xx is xx77. The derivative of xx88 with respect to xx is yy00.
  3. Differentiating Terms: Now we write down the differentiated equation:\newline1dydx+1+2ydydx+15y2dydx+6x=0-1\frac{dy}{dx} + 1 + 2y\frac{dy}{dx} + 15y^{2}\frac{dy}{dx} + 6x = 0
  4. Writing Differentiated Equation: Next, we collect all the terms involving dydx\frac{dy}{dx} on one side and the terms not involving dydx\frac{dy}{dx} on the other side:\newline\(-1\frac{dy}{dx} + 22y\frac{dy}{dx} + 1515y^{22}\frac{dy}{dx} = 1-1 - 66x
  5. Solving for (dydx):</b>Wefactorout$(dydx)(\frac{dy}{dx}):</b> We factor out \$(\frac{dy}{dx}) from the left side of the equation:\newline(\frac{dy}{dx})(\(-1 + 22y + 1515y^{22}) = 1-1 - 66x
  6. Solving for (dydx):</b>Wefactorout$(dydx)(\frac{dy}{dx}):</b> We factor out \$(\frac{dy}{dx}) from the left side of the equation:\newline(dydx)(1+2y+15y2)=16x(\frac{dy}{dx})(-1 + 2y + 15y^{2}) = -1 - 6xNow we solve for (dydx)(\frac{dy}{dx}) by dividing both sides of the equation by (1+2y+15y2):(-1 + 2y + 15y^{2}):\newline(dydx)=16x1+2y+15y2(\frac{dy}{dx}) = \frac{-1 - 6x}{-1 + 2y + 15y^{2}}

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