Find dy/dx by implicit differentiation. sqrt(3 x+y)=2+x^2y^2?
(1/2)(3x+y)^-1/2*(3 + dy/dx) = 2x*y^2 + 2 x^2 y*dy/dx
Rearrange to solve for dy/dx.
To find dy/dx by implicit differentiation, we need to differentiate both sides of the equation with respect to x, treating y as a function of x. Let's go step by step:
First, let's differentiate the left-hand side of the equation, sqrt(3x + y), with respect to x using the chain rule:
d/dx (sqrt(3x + y)) = (1/2)*(3x + y)^(-1/2)*(3 + dy/dx)
Next, let's differentiate the right-hand side of the equation, 2 + x^2y^2, with respect to x using the product rule:
d/dx (2 + x^2y^2) = 2(x^2y^2)' + (x^2)'(y^2)
= 2(2xy^2 + x^2(2y)(dy/dx)) + 2xy^2
Now that we have both derivatives, we can equate them:
(1/2)*(3x + y)^(-1/2)*(3 + dy/dx) = 2(2xy^2 + x^2(2y)(dy/dx)) + 2xy^2
Simplifying this equation, we can then solve for dy/dx:
(1/2)*(3 + dy/dx)/(sqrt(3x + y)) = 2(2xy^2 + 2x^3y(dy/dx)) + 2xy^2
[(3 + dy/dx)/(2sqrt(3x + y))] = 4xy^2 + 4x^3y(dy/dx) + 2xy^2
[(3 + dy/dx)/(2sqrt(3x + y))] - 4xy^2 = 4x^3y(dy/dx) + 2xy^2
Now, let's isolate dy/dx terms on one side and simplify:
[(3 + dy/dx)/(2sqrt(3x + y))] - 4xy^2 = 4x^3y(dy/dx) + 2xy^2
[(3 + dy/dx)/(2sqrt(3x + y))] - 4xy^2 - 2xy^2 = 4x^3y(dy/dx)
[(3 + dy/dx)/(2sqrt(3x + y))] - 6xy^2 = 4x^3y(dy/dx)
(3 + dy/dx)/(2sqrt(3x + y)) - 6xy^2 = 4x^3y(dy/dx)
[(3 + dy/dx) - 12xy^2sqrt(3x + y)]/(2sqrt(3x + y)) = 4x^3y(dy/dx)
Finally, we can solve for dy/dx by isolating it:
(3 + dy/dx) - 12xy^2sqrt(3x + y) = 8x^3y(dy/dx)*sqrt(3x + y)
3 + dy/dx - 12xy^2sqrt(3x + y) = 8x^3y(dy/dx)*sqrt(3x + y)
3 - 12xy^2sqrt(3x + y) = 8x^3y(dy/dx)*sqrt(3x + y) - dy/dx
3 - 12xy^2sqrt(3x + y) = dy/dx(8x^3y*sqrt(3x + y) - 1)
dy/dx = (3 - 12xy^2sqrt(3x + y))/(8x^3y*sqrt(3x + y) - 1)
Therefore, the derivative dy/dx by implicit differentiation of the given equation sqrt(3x + y) = 2 + x^2y^2 is (3 - 12xy^2sqrt(3x + y))/(8x^3y*sqrt(3x + y) - 1).