Answer :
y cos x = x^2 + y^2
Taking derivative:
y' cos x - y sin x = 2x + 2y y'
Simplifying:
y' = (- 2x - y sin x)/(2y - cos x)
Taking derivative:
y' cos x - y sin x = 2x + 2y y'
Simplifying:
y' = (- 2x - y sin x)/(2y - cos x)
Answer:
Step-by-step explanation:
Given is an implicit function of y in x.
[tex]ycos x = x^2+y^2\\[/tex]
The steps shown were the differentiation on both the sides
Left side done using product rule as well as chain rule and right side addition rule and chain rule
The steps got as
[tex]((y)(-sinx)+(cosx)(dy/dx)=2x+2y(dy/dx)[/tex] is correct
Next step is to group all dy/dx terms together
[tex]cosx(dy/dx) - 2y(dy/dx) = ysinx+2x[/tex] is also right step
Now rewrite left side as
[tex]\frac{dy}{dx} (cosx-2y)=2x+ysinx\\\frac{dy}{dx}=\frac{ysinx+2x}{cosx-2y}[/tex]
Thus both the steps and answer are correct.