find dy/dx assuming that y is differentiable with respect to x and x is also differentiable with respect to y.
a)xy+x^2y^2=x^3y^3
b)xsiny+ycosx=xy
use implicit differentiation.
Remember the product rule and the chain rule:
xy+x^2y^2=x^3y^3
y + xy' + 2xy^2 + 2x^2yy' = 3x^2y^3 + 3x^3y^2y'
y'(x + 2x^2y - 3x^3y^2) = 3x^2y^3-y-2xy^2
y' =
3x^2y^3-y-2xy^2
------------------------
x + 2x^2y - 3x^3y^2
do the other the same way:
xsiny+ycosx=xy
siny + xcosyy' - sinx*y + cosx*y' = y + xy'
...
can you add one more step am not just sure
collect all the y' stuff on one side
y'(x cosy+cosx-x) = y-siny-y sinx
Now just divide to get y' all by itself.
Don't forget your Algebra I now that you're taking calculus!
Don't clear me
a) To find dy/dx, we can use implicit differentiation. Let's differentiate both sides of the equation with respect to x:
d/dx(xy + x^2y^2) = d/dx(x^3y^3)
Using the product rule on the first term, we get:
y + x(dy/dx) + 2xy^2 + 2x^2y(dy/dx) = 3x^2y^3 + 3x^3y^2(dy/dx)
Rearranging the terms, we have:
y + 2xy^2 = 3x^2y^3 - x(dy/dx) + 3x^3y^2(dy/dx) - 2x^2y(dy/dx)
Now, let's isolate dy/dx:
dy/dx(-x - 2x^2y + 3x^3y^2) = y + 2xy^2 - 3x^2y^3
dy/dx = (y + 2xy^2 - 3x^2y^3) / (-x - 2x^2y + 3x^3y^2)
b) We can apply implicit differentiation to find dy/dx for this equation as well. Differentiating both sides with respect to x:
d/dx(xsiny + ycosx) = d/dx(xy)
Using the product rule on the first term, we get:
siny + x(dy/dx)cosy - sinx + y(dy/dx)cosx = y + x(dy/dx)
Rearranging the terms, we have:
siny - sinx = x(dy/dx) - y(dy/dx)cosx + x(dy/dx)cosy - y
Now, let's isolate dy/dx:
dy/dx(x - ycosx + xcosy) = siny - sinx - y
dy/dx = (siny - sinx - y) / (x - ycosx + xcosy)
To find dy/dx, we can use implicit differentiation. Implicit differentiation allows us to find the derivative of a function, even if it is not expressed explicitly in terms of x and y.
a) xy + x^2y^2 = x^3y^3
To differentiate both sides with respect to x, we can treat y as a function of x and use the product rule and chain rule.
Differentiating the left side:
d/dx(xy) = x(dy/dx) + y
Differentiating the right side:
d/dx(x^3y^3) = 3x^2y^3(dx/dx) + 3x^3y^2(dy/dx)
Combining the differentiated terms:
x(dy/dx) + y + 2x^2y^2(dy/dx) = 3x^2y^3 + 3x^3y^2(dy/dx)
Rearranging terms and isolating dy/dx:
(x + 2x^2y^2 - 3x^3y^2)dy/dx = 3x^2y^3 - y
dy/dx = (3x^2y^3 - y) / (x + 2x^2y^2 - 3x^3y^2)
Therefore, dy/dx for the given equation is (3x^2y^3 - y) / (x + 2x^2y^2 - 3x^3y^2).
b) xsiny + ycosx = xy
Following the same steps as above, we differentiate both sides with respect to x:
Differentiating the left side:
d/dx(xsiny) = x(dy/dx)cosy + (sin y)(dx/dx)
Differentiating the right side:
d/dx(xy) = y(dy/dx) + x
Combining the differentiated terms:
x(dy/dx)cosy + (sin y) + y(dy/dx) = y(dy/dx) + x
Rearranging terms and isolating dy/dx:
(x(dy/dx)cosy - y(dy/dx)) = x - (sin y)
(dy/dx)(xcosy - y) = x - sin y
dy/dx = (x - sin y) / (xcosy - y)
Therefore, dy/dx for the given equation is (x - sin y) / (xcosy - y).